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NCERT Class 11 Physics Chapter 10 Thermal Properties of Matter
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Thermal Properties of Matter
Chapter: 10
| Part – II |
EXERCISE
1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Ans: To convert the temperatures from Kelvin (K) to Celsius (°C) and Fahrenheit (°F):
°C = K – 273.15
°F = (K – 273.15) × 9/5 + 32
For Neon:
K = 24.57 K
°C = 24.57 – 273.15 = -248.58 °C
°F = (-248.58) × 9/5 + 32 = -415.44 °F
For Carbon Dioxide:
K = 216.55 K
°C = 216.55 – 273.15 = -56.60 °C
°F = (-56.60) × 9/5 + 32 = -69.88 °F
So, the triple points of Neon and Carbon Dioxide are:
Neon: -248.58 °C, -415.44 °F
Carbon Dioxide: -56.60 °C, -69.88 °F
2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?
Ans: The triple point of water in both scales:
In scale A: 200 A.
In scale B : 350 B.
The relationship between the two scales can be expressed using a linear transformation. Generally, if a temperature TA in scale A is related to TB in scale B by:
TA = mTB + c
Where m is the conversion factor and c is the offset.
TA = 200 A
TB = 350 B
We can write the relationship as:
200 A = m × 350 B + c
0A = m x 0B + c
Which implies c = 0 Thus,
200 A = m × 350 B
m = 200/350 = 4/7
Therefore, the relation between the temperatures TA and TB is:
TA = 4/7 TB.
3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = Ro [1 + α (T – To )] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
Ans: R = Ro [1 + α (T – To)]
We know two sets of values:
1. At the triple point of water (T = 273.16 K):
R = 101.6 Ω
2. At the normal melting point of lead (T = 600.5 K):
R = 165.5 Ω
We need to determine α and R0 first.
1. 101.6 = Ro [1 + α (273.16 – To)]
2. 165.5 = Ro [1 + α (600.5 – To)]
Now, divide equation (2) by equation (1):
(165.5 / 101.6) = [1 + α (600.5 – To)] / [1 + α (273.16 – To)]
α ≈ 0.0025 K⁻¹
Now that we have α, substitute it back into one of the original equations to find Ro:
Ro ≈ 100 Ω
Now that we have Ro and α, we can find the temperature when
R = 123.4 Ω:
123.4 = 100 [1 + 0.0025 (T – To)]
Solve for T:
T = 373.2 K
So, the temperature when the resistance is 123.4 Ω is approximately 373.2 K.
4. Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
Ans: Triple point of water has a unique value, i.e. 273.16 k. The temperature of the triple point of water is a standard fixed point because it is unique and reproducible, regardless of pressure changes.
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
Ans: The other fixed point is absolute zero of temperature.
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T – 273.15 Why do we have 273.15 in this relation, and not 273.16?
Ans: The temperature 273.16 K is the triple point of water. 273.15 K is used because it aligns with the historical Celsius scale definition, not because it is the melting point of ice.
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
Ans: Tc = 273.16 – 273.15
= 0.01°C
Celsius temperature to Fahrenheit:
tf = 9/5 × tc + 32
tf = 9/5 × 0.01 + 32
tf = 0.018 + 32 = 32.018 °F.
Convert this Fahrenheit temperature to the new absolute scale:
TR = tf + 459.67
TR = 32.018 + 459. 67
= 491.688 R.
5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:
| Temperature | Pressure thermometer A | Pressure thermometer B |
| Triple-point of water. | 1.250 x 105 Pa | 0.200 x 105 Pa |
| Normal melting point of sulphur. | 1.797 x 105 Pa | 0.287 x 105 Pa |
(a) What is the absolute temperature of the normal melting point of sulphur as read by thermometers A and B?
Ans: For thermometer A
Ptr = 1.250 × 105 Pa
Ttr = 273.16 k
P = 1.797 × 105 Pa and T=?
Now, using, Ptr / Ttr
= P/T.
T =
= 1.797 × 105 × 273.16 / 1.250 × 105
= 392. 69 K.
For thermometer B,
Ptr = 0.200 × 103 Pa
Ttr = 273.16 k
P = 0.287 × 105
T = ?
T =
= 2.287 × 105 × 273.16 / 0.200 × 105
= 391.98k.
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Ans: The discrepancy between the two readings is due to the fact that the gases are not perfectly ideal gases. To reduce the discrepancy, the reading should be taken at low pressures, so that the gases could show perfect behaviour:
For a temperature of triple point, i.e., K, the temperature on the new scale is:
= 273.16 × 180/100
= 491.688.
6. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1.
Ans: Given:
Measured length = 63.0 cm.
Temperature when measured = 45.0°C
Calibration temperature = 27.0 °C
Coefficient of linear expansion (α) = 1.20 × 10-5 k-1
The actual length (Lactual) is given by:
Lactual = Lmeasured / 1 + α × (Lmeasured – Tcalibrated)
Substitute the values:
Lactual = 63.0 cm / 1 + 1.20 × 10-5 × (45.0 – 27.0)
Lactual = 63.0cm / 1 + 1.20 × 10-5 × 18.0
Lactual = 63.0cm / 1+ 2.16 × 10-4
Lactual = 63.0cm / 1.000216
= 62.98 cm.
The length of the steel rod at 27.0°C is:
63.0 cm.
7. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10–5 K–1.
Ans: Initial differences
Δd = 8.70 cm – 8.69 cm = 0.01 cm
Δd = α × d × ΔT
where d is the initial diameter of the shaft (8.70 cm),
α is the coefficient of linear expansion (1.20 x 10–5 K–1),
and ΔT is the change in temperature.
Formula to solve for ΔT:
ΔT = Δd / (α × d)
= 0.01 cm / (1.20 × 10–5 K–1 × 8.70 cm)
= 96.3 K
Since the shaft is cooled from 27 °C, the final temperature (T) will be:
T = 27 °C – 96.3 K
= -69.3 °C
8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1 .
Ans: Here, Δd = α × d × ΔT
where:
α = 1.70 × 10–5 K–1
d = 4.24 cm
ΔT = 227 °C – 27 °C = 200 °C
First, convert ΔT to Kelvin:
ΔT = 200 °C = 200 K
Δd = 1.70 × 10–5 K–1 × 4.24 cm × 200 K
= 0.0144 cm
So, the diameter of the hole increases by 0.0144 cm when the sheet is heated to 227 °C.
The new diameter will be:
4.24 cm + 0.0144 cm = 4.2544 cm.
9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.
Ans: Calculate the thermal strain:
Thermal strain = α × ΔT
= 2.0 × 10-5 K-1 × (-66K) = -1.32 × 10-3
Calculate the tension developed:
= ΔL/L = F / Y × A
Give that ΔL / L = – α ΔT
– α ΔT = F / Y × A
Rearranging formula to find F:
F = – (α ΔT) × Y × A
Calculate the cross-sectional area (A) of the wire:
A = 𝜋 × (d/2)2
Where d is the diameter of the wire:
A = 𝜋 × (2.0mm / 2)2 = 𝜋 × (1.0mm)2 = 𝜋 (1.0 × 10-3 m)2 = 𝜋 × 10-6 m2
Substitute the values to find the tension F:
F = – (-1.32 × 10-3) × (0.91 × 1011 Pa) × (𝜋 × 10-6 m2)
F = 1.32 × 10-3 x 0.91 × 1011 × 𝜋 × 10-6 N
F = 1.32 × 0.91 × 𝜋 × 102 N
F = 3.77 × 102 N
= 377 N.
10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 x 10–5 K–1, steel = 1.2 x 10–5 K–1).
Ans: Calculating the change in length:
ΔL = α × L × ΔT where:
ΔL is the change in length
α is the coefficient of linear expansion
L is the original length
ΔT is the change in temperature
For Brass:
ΔLbrass = 2.0 × 10-5 K-1 × 0.5 m × (250 °C – 40 °C)
ΔLbrass = 2.1 × 10-3 m = 2.1 mm
For Steel:
ΔLsteel = 1.2 × 10-5 K-1 × 0.5 m × (250 °C – 40 °C)
ΔLsteel = 1.26 × 10-3 m = 1.26 mm
Total Change in Length:
ΔLtotal = ΔLbrass + ΔLsteel
ΔLtotal = 2.1 mm + 1.26 mm
ΔLtotal = 3.36 mm.
11. The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature?
Ans: Here, r, = 49 × 10-5 °C-1
ΔT = 30 °C
Fractional change,
Δp/ p = yΔT
= 49 × 10-5 × 30 = 1.47 × 10-2.
12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1 K–1 .
Ans: Power used for heating the block:
= 10 kW × 0.5 = 5 kW = 5000 W
Convert time into second:
Time = 2.5 minutes = 150 seconds
Energy transferred
= Power × Time
= 5000 W × 150 s = 750,000 J
Mass of the block
= 8.0 kg = 8000 g
Specific heat of aluminium = 0.91 J/g°C
formula: = mcΔT
750,000 J = 8000 g × 0.91 J/g°C × ΔT
ΔT = 750,000 J / (8000 g × 0.91 J/g°C)
= 103.3°C.
13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1).
Ans: Q = m × c × ΔT
m = 2.5 kg × 1000g/kg = 2500g
Q = 2500g × 0.39 J /g °C (500 °C – 0 °C)
Q = 2500 × 0.39 × 500
Q = 2500 × 195
Q = 487500 J
Calculate the amount of ice that can be melted:
mice = Q/ Lf
mice = 487500J / 335 j/g
mice = 1457.46 g
mice = 1.46 kg.
14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Ans: Mass of the metal, m = 0.20 kg = 200g
Initial temperature of the metal, T1 = 150°C
Final temperature of the metal, T2 = 40 °C
Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C
150 x 1 = 150g
Fall in the temperature of the metal:
ΔT = T1 – T2 = 150 – 40 = 110 °C
Specific heat of the water, Cw = 4.186 j/g °K
ΔT = 40 – 27 = 13°C
mCΔT = (M + m ) CwΔT
200 x C x 110 = (150 + 25) x 4.186 x 13
C = 175 x 4.186 x 13 / 110 x 200
= 0.43 Jg-1 K-1.
15. Given below are observations on molar specific heats at room temperature of some common gases.
| Gas | Molar specific heat (Cv) (cal mol-1 k-1) |
| Hydrogen | 4.87 |
| Nitrogen | 4.97 |
| Oxygen | 5.02 |
| Nitric oxide | 4.99 |
| Carbon monoxide | 5.01 |
| Chlorine | 6.17 |
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
Ans: Monatomic Gases, For monatomic gases, the molar specific heat at constant volume is typically around 2.92 cal/mol K. This value is based on the fact that monatomic gases have only translational degrees of freedom.
The measured molar sp. The heats of the gases mentioned in the table are markedly different because all these gases are diatomic gases.
A monatomic gas has only translational motion. In a diatomic gas, apart from translational motion, vibrational as well as rotational motion is also possible.
Therefore, to increase the temperature of these gases through 1°C, an extra amount of heat is Required to increase the average energy of different molecules in motion. Under ordinary temperature conditions (say room temperature) diatomic gases usually have rotational motion apart from their translational motion. But, for chlorine, vibrational motion also occurs. This is the reason that chlorine has a somewhat large value of molar specific heat.
16. A child running a temperature of 101°F is given an antipyrine (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1 .
Ans: Initial temperature of the child, T1 = 101°F
Final temperature of the child, T2 = 98 °F
Time taken to reduce the fever = 20 minutes
Mass of the child, m = 30 kg
Specific heat of the human body, c = 1 cal/g. °C
Latent heat of evaporation of water, L = 580 cal/g
Temperatures to celsius:
T1 = 5/9 x (101 – 32) = 5/9 x 69
= 38.33 °C.
T2 = 5/9 x (98 – 32) = 5/9 x 66
= 36.67 °C
Heat lost by body:
Q = m.c.ΔT
Where:
ΔT = T1 – T2 = 38.33 – 36 . 67
= 1.66 °C
m = 30 x 1000 = 30000g
Q = 30000g x 1 cal/g °C x 1.66 °C
= 49800 cal.
Sweat evaporated:
Q = msweet x L
Where L = 580 cal/g
Solving for msweat:
msweat = Q/L
= 49800 cal / 580 cal/g
= 85.86g
Rate of evaporation:
msweat / time
= 85.86g / 20 minutes
= 4.29g/min.
17. A ‘thermocol’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and the coefficient of thermal conductivity of thermocol is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1].
Ans: Surface area = 6 x (0.3 m)² = 0.54 m²
Calculate the heat flow through the thermocol walls:
= (0.01 J s⁻¹ m⁻¹ K⁻¹) x (0.54 m²) x (45°C – 0°C) / (0.05 m)
= 48.6 J/s
Hotel Heat in 6 hours:
Total heat flow = Heat flow x Time
= 48.6 J/s x (6 h x 3600 s/h)
= 835,040 J
Calculate the mass of ice melted:
= 835,040 J / (335 x 10³ J/kg)
= 2.49 kg
Initial mass of ice = 4.0 kg
Mass of ice melted = 2.49 kg
Ice remaining = Initial mass of ice – Mass of ice melted
= 4.0 kg – 2.49 kg
= 1.51 kg
18. A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s–1 m–1 K–1 ; Heat of evaporation of water = 2256 x 103 J kg–1 .
Ans: A = 0.15 m2
d = 1.0 cm = 0.01m
m = 6.0 kg/min
k = 109 js-1 m-1 k-1
L = 2256 x 103 j/kg
Heat require:
Q =m x L
Rate of boiling = 6.0 kg / 60s
= 0.1 kg/s
Q = 0.1 kg/s x 2256 x 103 j/kg
= 225600 j/s
Temperature difference:
Q = k x A x ΔT / d
Rearrange to solve for ΔT:
ΔT = Q x d / k x A
ΔT = 225600 J/s x 0.01m / 109 J s-1m-1k-1 x 0.15m2
ΔT = 2256/ 16.35
ΔT = 137. 5
Temperature of the flame:
Tflame = 100°C + ΔT
Tflame = 100°C + 137.5 °C
Tflame = 237.5 °C .
19. Explain why :
(a) A body with large reflectivity is a poor emitter.
Ans: A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiation. This is because a body with large reflectivity is a poor absorber of heat, and poor absorbers of heat are poor emitters.
(b) A brass tumbler feels much colder than a wooden tray on a chilly day.
Ans: Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler.
(c) An optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
Ans: The temperature of red hot iron in the oven is given by E1 = 𝛔 sigma T4 When iron is taken out in the open temperature (T0) its radiation energy is given by, Es = 𝛔 (T4 – T04). Thus, the pyrometer measures. low values for the red hot iron in the open.An optical pyrometer measures temperature based on the intensity of radiation from an object. It is calibrated assuming the object is an ideal black body.
(d) The earth without its atmosphere would be inhospitably cold.
Ans: If the atmosphere of earth were not there, its surface would become too cold to live.
(e) Heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
Ans: Steam at 100∘ possesses more heat than the same mass of water at 100∘. One gram of steam at 100∘ possesses 540 calories of heat more than that possesses by 1 gm of water at 100∘.
20. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
Ans: Newton’s Law of cooling:
T (t) = Ts + (T0 – Ts) e-kt
(T0 is the initial temperature)
T1 = Ts + (Tinitial – Ts)e-kt1
= 50 = 20 + (80 – 20) e -5k
50 = 20 + 60e-5k
30 = 60e-5k
e-5k = 30/60 = ½
Second cooling interval:
T2 = Ts + (Tinitial – Ts)e-kt2
30 = 20 + (60 – 20) e-kt2
30 = 20 + 40 e-kt2
e-kt2 = ¼
Exponential terms:
e-kt2 = (e-5k) t2/t1
= ¼ = (½) t2/5
Solve for t2
= ¼ = (½) t2/5
= ¼ = (½)2
= t2/5 = 2
= t2 = 10 minutes.
