Balbharti Maharashtra State Board 11th Physics Textbook Solutions
Chapter 7 Thermal Properties of Matter Textbook Exercise Questions and Answers.
1. Choose the correct option.
Question 1.
Range of temperature in a clinical thermometer, which measures the temperature of human body, is
(A) 70 ºC to 100 ºC
(B) 34 ºC to 42 ºC
(C) 0 ºF to 100 ºF
(D) 34 ºF to 80 ºF
Answer:
(B) 34 ºC to 42 ºC
Question 2.
A glass bottle completely filled with water is kept in the freezer. Why does it crack?
(A) Bottle gets contracted
(B) Bottle is expanded
(C) Water expands on freezing
(D) Water contracts on freezing
Answer:
(C) Water expands on freezing
Question 3.
If two temperatures differ by 25 °C on Celsius scale, the difference in temperature on Fahrenheit scale is
(A) 65°
(B) 45°
(C) 38°
(D) 25°
Answer:
(B) 45°
Question 4.
If α, β and γ are coefficients of linear, area l and volume expansion of a solid then
(A) α: β:γ 1:3:2
(B) α:β:γ 1:2:3
(C) α:β:γ 2:3:1
(D) α:β:γ 3:1:2
Answer:
(B) α:β:γ 1:2:3
Question 5.
Consider the following statements-
(I) The coefficient of linear expansion has dimension K-1
(II) The coefficient of volume expansion has dimension K-1
(A) I and II are both correct
(B) I is correct but II is wrong
(C) II is correct but I is wrong
(D) I and II are both wrong
Answer:
(A) I and II are both correct
Question 6.
Water falls from a height of 200 m. What is the difference in temperature between the water at the top and bottom of a water fall given that specific heat of water is 4200 J kg-1 °C-1?
(A) 0.96 °C
(B) 1.02 °C
(C) 0.46 °C
(D) 1.16 °C
Answer:
(C) 0.46 °C
2. Answer the following questions.
Question 1.
Clearly state the difference between heat and temperature?
Answer:
Question 2.
How a thermometer is calibrated?
Answer:
Question 3.
What are different scales of temperature? What is the relation between them?
Answer:
Relation between different scales of temperature:
\(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}=\frac{\mathrm{T}_{\mathrm{C}}-0}{100}=\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\)
where,
TF = temperature in fahrenheit scale,
TC = temperature in celsius scale,
TK = temperature in kelvin scale,
[Note: At zero of the kelvin scale, every substance in nature has the least possible activity.]
Question 4.
What is absolute zero?
Answer:
Question 5.
Derive the relation between three coefficients of thermal expansion.
Answer:
Consider a square plate of side l0 at 0 °C and h at T °C.
Relation between coefficient of linear expansion (α) and coefficient of cubical expansion (γ).
Relation between α, β and γ is given by,
α = \(\frac{\beta}{2}=\frac{\gamma}{3}\)
where, α = coefficient of linear expansion.
β = coefficient of superficial expansion,
γ = coefficient of cubical expansion.
Question 6.
State applications of thermal expansion.
Answer:
Applications of thermal expansion:
Question 7.
Why do we generally consider two specific heats for a gas?
Answer:
Question 8.
Are freezing point and melting point same with respect to change of state ? Comment.
Answer:
Though freezing point and melting point mark same temperature (0°C or 32° F), state of change is different for the two points. At freezing point liquid gets converted into solid, whereas at melting point solid gets converted into its liquid state.
Question 9.
Define
(i) Sublimation
(ii) Triple point.
Answer:
Question 10.
Explain the term ‘steady state’.
Answer:
Question 11.
Define coefficient of thermal conductivity. Derive its expression.
Answer:
Coefficient of thermal conductivity of a material is defined as the quantity of heat that flows in one second between the opposite faces of a cube of side 1 m, the faces being kept at a temperature difference of 1°C (or 1 K).
Expression for coefficient of thermal conductivity:
Question 12.
Give any four applications of thermal conductivity in every day life.
Answer:
Answer: Applications of thermal conductivity:
Question 13.
Explain the term thermal resistance. State its SI unit and dimensions.
Answer:
The SI unit of thermal resistance is °C s/kcal or °C s/J and its dimensional formula is [L-2M-1T3K1].
Question 14.
How heat transfer occurs through radiation in absence of a medium?
Answer:
Question 15.
State Newton’s law of cooling and explain how it can be experimentally verified.
Answer:
The rate of loss of heat dT/dt of the both’ is directly proportional to the difference of temperature (T – T0) of the body and the surroundings provided the difference in temperatures is small.
Mathematically, Newton’s law of cooling can be expressed as:
\(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ (T – T0)
∴ \(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ C(T – T0)
where, C is constant of proportionality. Experimental verification of Newton’s law of cooling:
Question 16.
What is thermal stress? Give an example of disadvantages of thermal stress in practical use?
Answer:
Question 17.
Which materials can be used as thermal insulators and why?
Answer:
3. Solve the following problems.
Question 1.
A glass flask has volume 1 × 10-4 m3. It is filled with a liquid at 30 ºC. If the temperature of the system is raised to 100 ºC, how much of the liquid will overflow. (Coefficient of volume expansion of glass is 1.2 × 10-5 (ºC)-1 while that of the liquid is 75 × 10-5 ºC-1)
Solution:
Given: V1 = 1 × 10-4 m3 = 10-4 m3, T1 = 30°C,
T2 = 100 °C
To find: Volume of liquid that overflows
Formula: γ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
Increase is volume = V2 – V1
= γV1(T2 – T1)
increase in volume of beaker
= γglass × V1 (T2 – T1)
= 1.2 × 10-5 × 10-4 × (100 – 30)
= 1.2 × 70 × 10-9
= 4 × 10-9 m3
∴ Increase in volume of beaker
= 84 × 10-9 m3
Increase in volume of liquid
= γliquid × V1 (T2 – T1)
= 75 × 10-5 × 10 × (100 – 30)
= 75 × 70 × 10
= 5250 × 10-9 m3
∴ Increase in volume of liquid = 5250 × 10-9 m3
∴ Volume of liquid which overflows
= (5250 – 84) × 10-9 m3
= 5166 × 10-9 m3
= 0.5166 × 10-7 m3
Volume of liquid that overflows is 0.5166 × 10-7 m3.
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]
Question 2.
Which will require more energy, heating a 2.0 kg block of lead by 30 K or heating a 4.0 kg block of copper by 5 K? (slead = 128 J kg-1 K-1, scopper = 387 J kg-1 K-1)
Solution:
Given: mlead = 2 kg, ∆Tlead = 30 K,
slead = 128 J/kg K,
mCu =4 kg, ∆TCu = 5 K,
sCu = 387 J/kg K
To find: Substance requiring more heat energy.
Formula: Q = ms ∆T
Calculation: From formula,
For lead, Qlead = 2 × 128 × 30 = 7680J
For Copper, QCu = 4 × 387 × 5 = 7740 J
QCu > Qlead, copper will require more heat energy.
Copper will require more heat energy.
Question 3.
Specific latent heat of vaporization of water is 2.26 × 106 J/kg. Calculate the energy needed to change 5.0 g of water into steam at 100 ºC.
Solution:
Given: Lvap = 2.26 × 106 J/kg
m = 5g = 5 × 10-3 kg
In this case, no temperature change takes place only change of state occurs.
To find: Heat required to convert water into steam.
Formula: Heat required = mLvap
Calculation: From formula,
Heat required = 5 × 10-3 × 2.26 × 106
= 11300J
= 1.13 × 104 J
Heat required to convert water into steam is 1.13 × 104 J
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]
Question 4.
A metal sphere cools at the rate of 0.05 ºC/s when its temperature is 70ºC and at the rate of 0.025 ºC/s when its temperature is 50 ºC. Determine the temperature of the surroundings and find the rate of cooling when the temperature of the metal sphere is 40 ºC.
Solution:
∴ 2(50 – T0) = 70 – T0
∴ T0 = 30 πC
Substituting value of T0.
0.05 = C (70 – 30)
∴ C = \(\frac{0.05}{40}\) = 0.00125/s.
For T3 = 40 °C
\(\left(\frac{\mathrm{d} \mathrm{T}}{\mathrm{dt}}\right)_{3}\) = C(T3 – T0)
= 0.00125 (40 – 30)
= 0.00125 × 10
= 0.0125°C/s.
i) Temperature of surrounding is 30 °C.
ii) Rate of cooling at 40 °C is 0.0125 °C/s.
Question 5.
The volume of a gas varied linearly with absolute temperature if its pressure is held constant. Suppose the gas does not liquefy even at very low temperatures, at what temperature the volume of the gas will be ideally zero?
Answer:
At temperature of -273.15 °C, the volume of the gas will be ideally zero.
Question 6.
In olden days, while laying the rails for trains, small gaps used to be left between the rail sections to allow for thermal expansion. Suppose the rails are laid at room temperature 27 ºC. If maximum temperature in the region is 45 ºC and the length of each rail section is 10 m, what should be the gap left given that α = 1.2 × 10-5 K-1 for the material of the rail section?
Solution:
Given. T1 = 27 °C, T2 = 45 °C,
L1 = 10m.
α = 1.2 × 10-5 K-1
To find: Gap that should be left (L2 – L1)
Formula: L2 – L1 = L1 α(T2 – T1)
Calculation: From formula,
L2 – L1 = 10 × 1.2 × 10-5 × (45 – 27)
= 2.16 × 10-3 m
= 2.16 mm
The gap that should be left between rail sections is 2.16 mm.
Question 7.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the wooden rim and the iron ring are 1.5 m and 1.47 m respectively at room temperature of 27 ºC. To what temperature the iron ring should be heated so that it can fit the rim of the wheel (αiron = 1.2 × 10-5 K-1).
Solution:
Given: dw = 1.5 m, d = 1.47 m, T1 = 27 °C.
αi = 1.2 × 10-5/ K
To find: Temperature (T2)
Formula. α = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}\)
Calculation: From formula,
T2 = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}} \alpha}\) + T1
= \(\frac{1.5-1.47}{1.47 \times 1.2 \times 10^{-5}}\) + 27
= 1700.7 + 27
= 1727.7 °C
Iron ring should be heated to temperature of 1727.7 °C.
Question 8.
In a random temperature scale X, water boils at 200 °X and freezes at 20 °X. Find the boiling point of a liquid in this scale if it boils at 62 °C.
Solution:
Here thermometric property P is temperature at some random scale X.
Using equation,
T = \(\frac{100\left(P_{T}-P_{1}\right)}{\left(P_{2}-P_{1}\right)}\)
For P1 = 20 °X,
P2 = 200 °X,
T = 62°C
∴ 62 = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-20\right)}{(200-20)}\)
∴ PT = \(\frac{62 \times(200-20)}{100}\) + 20 = 111.6 + 20
= 131.6 °X
The boiling point of a liquid in this scale is 131.6 °X.
Question 9.
A gas at 900°C is cooled until both its pressure and volume are halved. Calculate its final temperature.
Solution:
Given: T1 = 900 °C = 900 + 273.15 = 1173.15 K
V2 = \(\frac{\mathrm{V}_{1}}{2}\), P2 = \(\frac{\mathrm{P}_{1}}{2}\)
To find: Final temperature (T2)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{\mathrm{I}}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula.
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{1173.15}=\frac{\mathrm{P}_{1} \mathrm{~V}_{\mathrm{l}}}{4 \mathrm{~T}_{2}}\)
∴ T2 = \(\frac{1173.15}{4}\) = 293.29 K
Final temperature of gas is 293.29 K.
Question 10.
An aluminium rod and iron rod show 1.5 m difference in their lengths when heated at all temperature. What are their lengths at 0 °C if coefficient of linear expansion for aluminium is 24.5 × 10-6 /°C and for iron is 11.9 × 10-6 /°C
Solution:
Given: (LT)i – (LT)al = 1.5 m, T0 = 0 °C
αal = 24.5 × 10-6/°C
αi = 11.9 × 10-6 /°C
To find: Lengths of aluminium and iron rod (L0)al and (L0)i
Formula: LT = L0[(1 + α(T – T0)]
Calculation: For T0 = 0 °C
From formula,
LT = L0(1 + αT)
For aluminium,
(L0)al = (L0)al(1 + αalT) ……………. (1)
For iron,
(LT)i = (L0)i (1 + αiT) ………….. (2)
Subtracting equation (2) by (1),
(LT)i – (LT)al = [(L0)i + (L0)i αiT] – [(L0)al + (L0)alαalT]
= (L0)i – (L0)al + [(L0)i αi – (L0)al αal]T
∴ 1.5 = 1.5 + [(L0)i αi – (L0)al αal)]T
⇒ [(L0)iαi – (L0)alαal] T = 0
∴ (L0)alαal = (L0)iαi
Length of aluminium rod at 0 °C is 1.417 m and that of iron rod is 2.917 m.
Question 11.
What is the specific heat of a metal if 50 cal of heat is needed to raise 6 kg of the metal from 20°C to 62 °C ?
Solution:
Given: Q = 50 cal, m =6 kg,
∆T = 62 – 20 = 42 °C
To find: Specific heat (s)
Formula: Q = ms ∆T
Calculation: From formula,
s = \(\frac{\mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}=\frac{50}{6 \times 42}\) = 0.198 cal/kg °C
Specific heat of metal is copper 0.198 cal/kg °C.
Question 12.
The rate of flow of heat through a copper rod with temperature difference 30 °C is 1500 cal/s. Find the thermal resistance of copper rod.
Solution:
Given: ∆T = 30 °C, Pcond = 1500 cal/s
To find: Thermal resistance (RT)
Formula: RT = \(\frac{\Delta \mathrm{T}}{\mathrm{P}_{\text {cond }}}\)
Calculation: From formula,
RT = \(\frac{30}{1500}\)
= 0.02 °C s/cal.
Thermal resistance of copper rod is 0.02 °C s/cal.
Question 13.
An electric kettle takes 20 minutes to heat a certain quantity of water from 0°C to its boiling point. It requires 90 minutes to turn all the water at 100°C into steam. Find the latent heat of vaporisation. (Specific heat of water = 1cal/g°C)
Solution:
Let heat supplied by kettle in 20 minutes be Q1 and that in 90 min. be Q2.
Using heat temperature of water is raised from O °C to 100 °C.
If mass of water in the kettle is ‘m’ then.
Q1 = mswater∆T m × 1 × (100 – 0)
= 100 m ………….. (i)
…………. (∵ Swater = 1 cal/g °C)
Similarly using heat Q2 water is converted from liquid to gas,
∴ Q2 = mLvap ……………. (ii)
Given that heat Q1, Q2 are supplied to water in 20 min. (t1) and 90 min (t2) respectively.
Kettle being same its conduction rate (Pcond) is same.
Using Pcond = \(\frac{\mathrm{Q}_{1}}{\mathrm{t}_{1}}=\frac{\mathrm{Q}_{2}}{\mathrm{t}_{2}}\) …………… (iii)
From (i), (ii) and (iii),
\(\frac{100 \mathrm{~m}}{20}=\frac{\mathrm{mL}_{\text {vap }}}{90}\),
∴ Lvap = 5 × 90 = 450 cal/g
Latent heat of vaporisation for water is 450 cal/g.
Question 14.
Find the temperature difference between two sides of a steel plate 4 cm thick, when heat is transmitted through the plate at the rate of 400 k cal per minute per square metre at steady state. Thermal conductivity of steel is 0.026 kcal/m s K.
Solution:
Temperature difference between two sides is 10.26 K.
[Note: Above answer is expressed in K (‘kelvin considering that thermal conductivity is expressed in units of kcal / ms K, and not as kcal / m s °C. As 1 °C equivalent to 1 K. conceptually temperature difference of 10.26 K will correspond to 10.26 t]
Question 15.
A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30°C?
Solution:
Given: T1 = 80 °C, T2 = 60 °C, T3 = 40 °C, T0 = 30 °C, (dt)1 = 6 min.
To find: Time taken in cooling (dt)2
Time taken in cooling is 10 min.
11th Physics Digest Chapter 7 Thermal Properties of Matter Intext Questions and Answers
Can you tell? (Textbook Page No. 125)
Question 1.
i) Why the metal wires for electrical transmission lines sag?
ii) Why a railway track is not a continuous piece but is made up of segments separated by gaps?
iii) How a steel wheel is mounted on an axle to fit exactly?
Answer:
Intext question. (Textbook Page No 124)
Question 1.
Can you now tell why the balloon bursts sometimes when you try to fill air in it?
Answer:
Can you tell? (Textbook Page No. 125)
Question 1.
Why lakes freeze first at the surface?
Answer:
Activity (Textbook Page No. 129)
Question 1.
To understand the process of change of state:
Take some cubes of ice in a beaker. Note the temperature of ice (0 °C). Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Observe the change in temperature. Continue heating even after the whole of ice gets converted into water. Observe the change in temperature as before till vapours start coming out. Plot the graph of temperature (along Y-axis) versus time (along X-axis). Obtain a graph of temperature versus time.
Answer:
[Students are expected to attempt the activity on their own.]
Can you tell? (Textbook Page No. 130)
Question 1.
What is observed after point D in graph? Can steam be hotter than 100 °C?
Answer:
Beyond point D, thermometer again shows rise in temperature. This means, steam can be hotter than 100 °C and is termed as superheated steam.
Question 2.
Why steam at 100 °C causes more harm to our skin than water at 100 °C?
Answer:
Activity (Textbook Page No. 130)
Activity to understand the dependence of boiling point on pressure:
Take a round bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask as shown in figure. As water in the flask gets heated, note that first the air, which was dissolved in the water comes out as small bubbles. Later bubbles of steam form at the bottom but as they rise to the cooler water near the top, they condense and disappear. Finally, as the temperature of the entire mass of the water reaches 100 oc, bubbles of steam reach the surface and boiling is said to occur. The steam in the flask may not be visible hut as it comes out of the flask, It condenses as tiny droplets of water giving a foggy appearance.
If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling starts again. Thus, boiling point increases with increase in pressure. Let us now remove the burner. Allow water to cool to about 80°C. Remove the thermometers and steam outlet. Close the flask with a air tight cork. Keep the flask turned upside down on a stand. Pour icecold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure and increases with increase in pressure.
Answer:
[Students are expected to attempt the activity an their own.]
Can you tell? (Textbook Page No. 131)
Question 1.
i) Why is cooking difficult at high altitude?
ii) Why is cooking faster in pressure cooker?
Answer:
Internet my friend (Textbook Page No. 139)
i) https ://hyperphysics. phy-astr.gsu.edul/base/hframe.html
ii) https://youtu.be/7ZKHc5J6R5Q
iii) https://physics. info/expansion
Answer:
[Students are expected to visit the above mentioned webs it es and collect more information about the thermal properties of matter.]
Chapter 7 Thermal Properties of Matter Textbook Exercise Questions and Answers.
1. Choose the correct option.
Question 1.
Range of temperature in a clinical thermometer, which measures the temperature of human body, is
(A) 70 ºC to 100 ºC
(B) 34 ºC to 42 ºC
(C) 0 ºF to 100 ºF
(D) 34 ºF to 80 ºF
Answer:
(B) 34 ºC to 42 ºC
Question 2.
A glass bottle completely filled with water is kept in the freezer. Why does it crack?
(A) Bottle gets contracted
(B) Bottle is expanded
(C) Water expands on freezing
(D) Water contracts on freezing
Answer:
(C) Water expands on freezing
Question 3.
If two temperatures differ by 25 °C on Celsius scale, the difference in temperature on Fahrenheit scale is
(A) 65°
(B) 45°
(C) 38°
(D) 25°
Answer:
(B) 45°
Question 4.
If α, β and γ are coefficients of linear, area l and volume expansion of a solid then
(A) α: β:γ 1:3:2
(B) α:β:γ 1:2:3
(C) α:β:γ 2:3:1
(D) α:β:γ 3:1:2
Answer:
(B) α:β:γ 1:2:3
Question 5.
Consider the following statements-
(I) The coefficient of linear expansion has dimension K-1
(II) The coefficient of volume expansion has dimension K-1
(A) I and II are both correct
(B) I is correct but II is wrong
(C) II is correct but I is wrong
(D) I and II are both wrong
Answer:
(A) I and II are both correct
Question 6.
Water falls from a height of 200 m. What is the difference in temperature between the water at the top and bottom of a water fall given that specific heat of water is 4200 J kg-1 °C-1?
(A) 0.96 °C
(B) 1.02 °C
(C) 0.46 °C
(D) 1.16 °C
Answer:
(C) 0.46 °C
2. Answer the following questions.
Question 1.
Clearly state the difference between heat and temperature?
Answer:
| Heat | Temperature | |
| i. | Heat is energy in transit. When two bodies at different temperatures are brought in contact, they exchange heat. OR Heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of their temperature difference. |
Temperature is a physical quantity that defines the thermodynamic state of a system. OR Heat transfer takes place between the body and the surrounding medium until the body and the surrounding medium are at the same temperature. |
| ii. | Heat exchange can be measured with the help of a calorimeter. | Temperature is measured with the help of a thermometer. |
| iii. | Heat (being a form of energy) is a derived quantity. | Temperature is a fundamental quantity. |
How a thermometer is calibrated?
Answer:
Question 3.
What are different scales of temperature? What is the relation between them?
Answer:
Relation between different scales of temperature:
\(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}=\frac{\mathrm{T}_{\mathrm{C}}-0}{100}=\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\)
where,
TF = temperature in fahrenheit scale,
TC = temperature in celsius scale,
TK = temperature in kelvin scale,
[Note: At zero of the kelvin scale, every substance in nature has the least possible activity.]
Question 4.
What is absolute zero?
Answer:
Question 5.
Derive the relation between three coefficients of thermal expansion.
Answer:
Consider a square plate of side l0 at 0 °C and h at T °C.
Relation between coefficient of linear expansion (α) and coefficient of cubical expansion (γ).
Relation between α, β and γ is given by,
α = \(\frac{\beta}{2}=\frac{\gamma}{3}\)
where, α = coefficient of linear expansion.
β = coefficient of superficial expansion,
γ = coefficient of cubical expansion.
Question 6.
State applications of thermal expansion.
Answer:
Applications of thermal expansion:
Question 7.
Why do we generally consider two specific heats for a gas?
Answer:
Question 8.
Are freezing point and melting point same with respect to change of state ? Comment.
Answer:
Though freezing point and melting point mark same temperature (0°C or 32° F), state of change is different for the two points. At freezing point liquid gets converted into solid, whereas at melting point solid gets converted into its liquid state.
Question 9.
Define
(i) Sublimation
(ii) Triple point.
Answer:
Question 10.
Explain the term ‘steady state’.
Answer:
Question 11.
Define coefficient of thermal conductivity. Derive its expression.
Answer:
Coefficient of thermal conductivity of a material is defined as the quantity of heat that flows in one second between the opposite faces of a cube of side 1 m, the faces being kept at a temperature difference of 1°C (or 1 K).
Expression for coefficient of thermal conductivity:
Question 12.
Give any four applications of thermal conductivity in every day life.
Answer:
Answer: Applications of thermal conductivity:
Question 13.
Explain the term thermal resistance. State its SI unit and dimensions.
Answer:
The SI unit of thermal resistance is °C s/kcal or °C s/J and its dimensional formula is [L-2M-1T3K1].
Question 14.
How heat transfer occurs through radiation in absence of a medium?
Answer:
Question 15.
State Newton’s law of cooling and explain how it can be experimentally verified.
Answer:
The rate of loss of heat dT/dt of the both’ is directly proportional to the difference of temperature (T – T0) of the body and the surroundings provided the difference in temperatures is small.
Mathematically, Newton’s law of cooling can be expressed as:
\(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ (T – T0)
∴ \(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ C(T – T0)
where, C is constant of proportionality. Experimental verification of Newton’s law of cooling:
Question 16.
What is thermal stress? Give an example of disadvantages of thermal stress in practical use?
Answer:
Question 17.
Which materials can be used as thermal insulators and why?
Answer:
3. Solve the following problems.
Question 1.
A glass flask has volume 1 × 10-4 m3. It is filled with a liquid at 30 ºC. If the temperature of the system is raised to 100 ºC, how much of the liquid will overflow. (Coefficient of volume expansion of glass is 1.2 × 10-5 (ºC)-1 while that of the liquid is 75 × 10-5 ºC-1)
Solution:
Given: V1 = 1 × 10-4 m3 = 10-4 m3, T1 = 30°C,
T2 = 100 °C
To find: Volume of liquid that overflows
Formula: γ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
Increase is volume = V2 – V1
= γV1(T2 – T1)
increase in volume of beaker
= γglass × V1 (T2 – T1)
= 1.2 × 10-5 × 10-4 × (100 – 30)
= 1.2 × 70 × 10-9
= 4 × 10-9 m3
∴ Increase in volume of beaker
= 84 × 10-9 m3
Increase in volume of liquid
= γliquid × V1 (T2 – T1)
= 75 × 10-5 × 10 × (100 – 30)
= 75 × 70 × 10
= 5250 × 10-9 m3
∴ Increase in volume of liquid = 5250 × 10-9 m3
∴ Volume of liquid which overflows
= (5250 – 84) × 10-9 m3
= 5166 × 10-9 m3
= 0.5166 × 10-7 m3
Volume of liquid that overflows is 0.5166 × 10-7 m3.
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]
Question 2.
Which will require more energy, heating a 2.0 kg block of lead by 30 K or heating a 4.0 kg block of copper by 5 K? (slead = 128 J kg-1 K-1, scopper = 387 J kg-1 K-1)
Solution:
Given: mlead = 2 kg, ∆Tlead = 30 K,
slead = 128 J/kg K,
mCu =4 kg, ∆TCu = 5 K,
sCu = 387 J/kg K
To find: Substance requiring more heat energy.
Formula: Q = ms ∆T
Calculation: From formula,
For lead, Qlead = 2 × 128 × 30 = 7680J
For Copper, QCu = 4 × 387 × 5 = 7740 J
QCu > Qlead, copper will require more heat energy.
Copper will require more heat energy.
Question 3.
Specific latent heat of vaporization of water is 2.26 × 106 J/kg. Calculate the energy needed to change 5.0 g of water into steam at 100 ºC.
Solution:
Given: Lvap = 2.26 × 106 J/kg
m = 5g = 5 × 10-3 kg
In this case, no temperature change takes place only change of state occurs.
To find: Heat required to convert water into steam.
Formula: Heat required = mLvap
Calculation: From formula,
Heat required = 5 × 10-3 × 2.26 × 106
= 11300J
= 1.13 × 104 J
Heat required to convert water into steam is 1.13 × 104 J
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]
Question 4.
A metal sphere cools at the rate of 0.05 ºC/s when its temperature is 70ºC and at the rate of 0.025 ºC/s when its temperature is 50 ºC. Determine the temperature of the surroundings and find the rate of cooling when the temperature of the metal sphere is 40 ºC.
Solution:
∴ 2(50 – T0) = 70 – T0
∴ T0 = 30 πC
Substituting value of T0.
0.05 = C (70 – 30)
∴ C = \(\frac{0.05}{40}\) = 0.00125/s.
For T3 = 40 °C
\(\left(\frac{\mathrm{d} \mathrm{T}}{\mathrm{dt}}\right)_{3}\) = C(T3 – T0)
= 0.00125 (40 – 30)
= 0.00125 × 10
= 0.0125°C/s.
i) Temperature of surrounding is 30 °C.
ii) Rate of cooling at 40 °C is 0.0125 °C/s.
Question 5.
The volume of a gas varied linearly with absolute temperature if its pressure is held constant. Suppose the gas does not liquefy even at very low temperatures, at what temperature the volume of the gas will be ideally zero?
Answer:
At temperature of -273.15 °C, the volume of the gas will be ideally zero.
Question 6.
In olden days, while laying the rails for trains, small gaps used to be left between the rail sections to allow for thermal expansion. Suppose the rails are laid at room temperature 27 ºC. If maximum temperature in the region is 45 ºC and the length of each rail section is 10 m, what should be the gap left given that α = 1.2 × 10-5 K-1 for the material of the rail section?
Solution:
Given. T1 = 27 °C, T2 = 45 °C,
L1 = 10m.
α = 1.2 × 10-5 K-1
To find: Gap that should be left (L2 – L1)
Formula: L2 – L1 = L1 α(T2 – T1)
Calculation: From formula,
L2 – L1 = 10 × 1.2 × 10-5 × (45 – 27)
= 2.16 × 10-3 m
= 2.16 mm
The gap that should be left between rail sections is 2.16 mm.
Question 7.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the wooden rim and the iron ring are 1.5 m and 1.47 m respectively at room temperature of 27 ºC. To what temperature the iron ring should be heated so that it can fit the rim of the wheel (αiron = 1.2 × 10-5 K-1).
Solution:
Given: dw = 1.5 m, d = 1.47 m, T1 = 27 °C.
αi = 1.2 × 10-5/ K
To find: Temperature (T2)
Formula. α = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}\)
Calculation: From formula,
T2 = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}} \alpha}\) + T1
= \(\frac{1.5-1.47}{1.47 \times 1.2 \times 10^{-5}}\) + 27
= 1700.7 + 27
= 1727.7 °C
Iron ring should be heated to temperature of 1727.7 °C.
Question 8.
In a random temperature scale X, water boils at 200 °X and freezes at 20 °X. Find the boiling point of a liquid in this scale if it boils at 62 °C.
Solution:
Here thermometric property P is temperature at some random scale X.
Using equation,
T = \(\frac{100\left(P_{T}-P_{1}\right)}{\left(P_{2}-P_{1}\right)}\)
For P1 = 20 °X,
P2 = 200 °X,
T = 62°C
∴ 62 = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-20\right)}{(200-20)}\)
∴ PT = \(\frac{62 \times(200-20)}{100}\) + 20 = 111.6 + 20
= 131.6 °X
The boiling point of a liquid in this scale is 131.6 °X.
Question 9.
A gas at 900°C is cooled until both its pressure and volume are halved. Calculate its final temperature.
Solution:
Given: T1 = 900 °C = 900 + 273.15 = 1173.15 K
V2 = \(\frac{\mathrm{V}_{1}}{2}\), P2 = \(\frac{\mathrm{P}_{1}}{2}\)
To find: Final temperature (T2)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{\mathrm{I}}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula.
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{1173.15}=\frac{\mathrm{P}_{1} \mathrm{~V}_{\mathrm{l}}}{4 \mathrm{~T}_{2}}\)
∴ T2 = \(\frac{1173.15}{4}\) = 293.29 K
Final temperature of gas is 293.29 K.
Question 10.
An aluminium rod and iron rod show 1.5 m difference in their lengths when heated at all temperature. What are their lengths at 0 °C if coefficient of linear expansion for aluminium is 24.5 × 10-6 /°C and for iron is 11.9 × 10-6 /°C
Solution:
Given: (LT)i – (LT)al = 1.5 m, T0 = 0 °C
αal = 24.5 × 10-6/°C
αi = 11.9 × 10-6 /°C
To find: Lengths of aluminium and iron rod (L0)al and (L0)i
Formula: LT = L0[(1 + α(T – T0)]
Calculation: For T0 = 0 °C
From formula,
LT = L0(1 + αT)
For aluminium,
(L0)al = (L0)al(1 + αalT) ……………. (1)
For iron,
(LT)i = (L0)i (1 + αiT) ………….. (2)
Subtracting equation (2) by (1),
(LT)i – (LT)al = [(L0)i + (L0)i αiT] – [(L0)al + (L0)alαalT]
= (L0)i – (L0)al + [(L0)i αi – (L0)al αal]T
∴ 1.5 = 1.5 + [(L0)i αi – (L0)al αal)]T
⇒ [(L0)iαi – (L0)alαal] T = 0
∴ (L0)alαal = (L0)iαi
Length of aluminium rod at 0 °C is 1.417 m and that of iron rod is 2.917 m.
Question 11.
What is the specific heat of a metal if 50 cal of heat is needed to raise 6 kg of the metal from 20°C to 62 °C ?
Solution:
Given: Q = 50 cal, m =6 kg,
∆T = 62 – 20 = 42 °C
To find: Specific heat (s)
Formula: Q = ms ∆T
Calculation: From formula,
s = \(\frac{\mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}=\frac{50}{6 \times 42}\) = 0.198 cal/kg °C
Specific heat of metal is copper 0.198 cal/kg °C.
Question 12.
The rate of flow of heat through a copper rod with temperature difference 30 °C is 1500 cal/s. Find the thermal resistance of copper rod.
Solution:
Given: ∆T = 30 °C, Pcond = 1500 cal/s
To find: Thermal resistance (RT)
Formula: RT = \(\frac{\Delta \mathrm{T}}{\mathrm{P}_{\text {cond }}}\)
Calculation: From formula,
RT = \(\frac{30}{1500}\)
= 0.02 °C s/cal.
Thermal resistance of copper rod is 0.02 °C s/cal.
Question 13.
An electric kettle takes 20 minutes to heat a certain quantity of water from 0°C to its boiling point. It requires 90 minutes to turn all the water at 100°C into steam. Find the latent heat of vaporisation. (Specific heat of water = 1cal/g°C)
Solution:
Let heat supplied by kettle in 20 minutes be Q1 and that in 90 min. be Q2.
Using heat temperature of water is raised from O °C to 100 °C.
If mass of water in the kettle is ‘m’ then.
Q1 = mswater∆T m × 1 × (100 – 0)
= 100 m ………….. (i)
…………. (∵ Swater = 1 cal/g °C)
Similarly using heat Q2 water is converted from liquid to gas,
∴ Q2 = mLvap ……………. (ii)
Given that heat Q1, Q2 are supplied to water in 20 min. (t1) and 90 min (t2) respectively.
Kettle being same its conduction rate (Pcond) is same.
Using Pcond = \(\frac{\mathrm{Q}_{1}}{\mathrm{t}_{1}}=\frac{\mathrm{Q}_{2}}{\mathrm{t}_{2}}\) …………… (iii)
From (i), (ii) and (iii),
\(\frac{100 \mathrm{~m}}{20}=\frac{\mathrm{mL}_{\text {vap }}}{90}\),
∴ Lvap = 5 × 90 = 450 cal/g
Latent heat of vaporisation for water is 450 cal/g.
Question 14.
Find the temperature difference between two sides of a steel plate 4 cm thick, when heat is transmitted through the plate at the rate of 400 k cal per minute per square metre at steady state. Thermal conductivity of steel is 0.026 kcal/m s K.
Solution:
Temperature difference between two sides is 10.26 K.
[Note: Above answer is expressed in K (‘kelvin considering that thermal conductivity is expressed in units of kcal / ms K, and not as kcal / m s °C. As 1 °C equivalent to 1 K. conceptually temperature difference of 10.26 K will correspond to 10.26 t]
Question 15.
A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30°C?
Solution:
Given: T1 = 80 °C, T2 = 60 °C, T3 = 40 °C, T0 = 30 °C, (dt)1 = 6 min.
To find: Time taken in cooling (dt)2
Time taken in cooling is 10 min.
11th Physics Digest Chapter 7 Thermal Properties of Matter Intext Questions and Answers
Can you tell? (Textbook Page No. 125)
Question 1.
i) Why the metal wires for electrical transmission lines sag?
ii) Why a railway track is not a continuous piece but is made up of segments separated by gaps?
iii) How a steel wheel is mounted on an axle to fit exactly?
Answer:
Intext question. (Textbook Page No 124)
Question 1.
Can you now tell why the balloon bursts sometimes when you try to fill air in it?
Answer:
Can you tell? (Textbook Page No. 125)
Question 1.
Why lakes freeze first at the surface?
Answer:
Activity (Textbook Page No. 129)
Question 1.
To understand the process of change of state:
Take some cubes of ice in a beaker. Note the temperature of ice (0 °C). Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Observe the change in temperature. Continue heating even after the whole of ice gets converted into water. Observe the change in temperature as before till vapours start coming out. Plot the graph of temperature (along Y-axis) versus time (along X-axis). Obtain a graph of temperature versus time.
Answer:
[Students are expected to attempt the activity on their own.]
Can you tell? (Textbook Page No. 130)
Question 1.
What is observed after point D in graph? Can steam be hotter than 100 °C?
Answer:
Beyond point D, thermometer again shows rise in temperature. This means, steam can be hotter than 100 °C and is termed as superheated steam.
Question 2.
Why steam at 100 °C causes more harm to our skin than water at 100 °C?
Answer:
Activity (Textbook Page No. 130)
Activity to understand the dependence of boiling point on pressure:
Take a round bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask as shown in figure. As water in the flask gets heated, note that first the air, which was dissolved in the water comes out as small bubbles. Later bubbles of steam form at the bottom but as they rise to the cooler water near the top, they condense and disappear. Finally, as the temperature of the entire mass of the water reaches 100 oc, bubbles of steam reach the surface and boiling is said to occur. The steam in the flask may not be visible hut as it comes out of the flask, It condenses as tiny droplets of water giving a foggy appearance.
If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling starts again. Thus, boiling point increases with increase in pressure. Let us now remove the burner. Allow water to cool to about 80°C. Remove the thermometers and steam outlet. Close the flask with a air tight cork. Keep the flask turned upside down on a stand. Pour icecold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure and increases with increase in pressure.
Answer:
[Students are expected to attempt the activity an their own.]
Can you tell? (Textbook Page No. 131)
Question 1.
i) Why is cooking difficult at high altitude?
ii) Why is cooking faster in pressure cooker?
Answer:
Internet my friend (Textbook Page No. 139)
i) https ://hyperphysics. phy-astr.gsu.edul/base/hframe.html
ii) https://youtu.be/7ZKHc5J6R5Q
iii) https://physics. info/expansion
Answer:
[Students are expected to visit the above mentioned webs it es and collect more information about the thermal properties of matter.]