Question 1

Q1. A triple point of water is 273.16 K. Express this temperature in Fahrenheit scale.

Accepted Answer

To express the triple point of water (273.16 K) in the Fahrenheit scale, we first convert Kelvin to Celsius and then Celsius to Fahrenheit.

Question 2

Q2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?

Accepted Answer

For any absolute temperature scale, the ratio of two temperatures is equal to the ratio of the corresponding values on that scale. The triple point of water is a standard reference point. In the Kelvin scale, the triple point of water is 273.16 K.

Question 3

Q3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R_0 [1 + α(T - T_0)]. The resistance is 3.0 Ω at the triple

Accepted Answer

We are given the relationship between resistance (R) and temperature (T) for a thermometer: R = R_0 [1 + α(T - T_0)]. We need to determine the constants R_0 and α using the given data points, and then use these constants to find the temperature corresponding to a resistance of 3.05 Ω.

Question 4

Q4. 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 or the boiling point of

Accepted Answer

Let's address each part of the question individually.

Question 5

Q5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:
Temperature          Pressure thermometer A          Pressure thermom

Accepted Answer

This question explores the working principles and limitations of ideal gas thermometers.

Question 6

Q6. 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 their origi

Accepted Answer

The change in length of the combined rod is 0.264 cm. No thermal stress is developed at the junction because the rods are free to expand.

Question 7

Q7. 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

Accepted Answer

The change in the diameter of the hole when the sheet is heated to 227 °C is 0.0144 cm.

Question 8

Q8. 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/minute when placed on a gas stove. Estimate the temperature of the part of the

Accepted Answer

The estimated temperature of the part of the flame in contact with the boiler is approximately 100.9 °C.

Question 9

Q9. Explain why:
(a) Two thin blankets put together are warmer than a single thick blanket of the same total thickness.
(b) A hot liquid cools more slowly when partly filled in a the

Accepted Answer

Here are the explanations for the given phenomena:

**(a) Two thin blankets put together are warmer than a single thick blanket of the same total thickness.**

**Explanation:** The warmth provided by blankets is primarily due to their ability to trap air. Air is a poor conductor of heat (a good insulator). When two thin blankets are put together, a layer of air is trapped between them, in addition to the air trapped within the fibers of each blanket. This additional layer of trapped air acts as an extra insulating barrier, significantly reducing heat loss from the body by conduction and convection. A single thick blanket, even of the same total thickness, might trap less air overall compared to the two separate blankets with an air gap between them, making the combined thin blankets more effective at preventing heat transfer.

**(b) A hot liquid cools more slowly when partly filled in a thermos flask than when it is completely filled.**

**Explanation:** A thermos flask is designed to minimize heat transfer by conduction, convection, and radiation. It has a double-walled vacuum space to prevent conduction and convection, and silvered surfaces to minimize radiation. However, heat loss can still occur through the stopper (conduction and convection) and through the exposed surface of the liquid (evaporation, convection, and radiation).

When the thermos flask is partly filled, there is a larger volume of air (or vapor) above the liquid. This air can facilitate heat loss through convection within the flask and then through the stopper. More importantly, the rate of cooling depends on the temperature difference between the liquid and its surroundings, and the surface area exposed for heat loss. While the surface area of the liquid might be smaller when partly filled, the primary reason for slower cooling is often related to the *mass* of the liquid. A completely filled flask has a larger mass of hot liquid. For a given rate of heat loss (which might be similar for both cases if the flask is well-insulated), a larger mass of liquid will experience a smaller drop in temperature over the same time period because of its higher thermal inertia (heat capacity). However, the question implies a comparison of cooling *rate* for the liquid itself. The most significant factor here is often the *convection* within the air/vapor space above the liquid. When partly filled, the air above the liquid can set up convection currents, transferring heat to the stopper more effectively than if the liquid itself were in direct contact with the stopper over a larger area. *Correction:* The more accurate explanation for slower cooling when partly filled is that the *rate of heat loss per unit mass* is lower. While the total heat loss might be similar (or even slightly higher due to convection in the air gap), the larger mass in a completely filled flask means its temperature will drop more slowly for the same amount of heat lost. However, the common understanding for thermos flasks is that the *air gap* above the liquid in a partly filled flask acts as an insulator, reducing heat loss through the stopper compared to when the hot liquid is in direct contact with the stopper over a larger area. Let's re-evaluate. The primary heat loss mechanism from the liquid surface in a thermos is evaporation and convection through the air/vapor space to the stopper. If the flask is partly filled, there's a larger volume of air/vapor above the liquid. This air/vapor itself can act as an insulating layer, reducing the direct contact and convective heat transfer from the liquid surface to the stopper compared to when the liquid level is higher and closer to the stopper. Also, the larger air column above the liquid means the heat has to travel a longer path through the air before reaching the stopper. This reduces the overall rate of heat transfer. Additionally, the specific heat capacity of the air/vapor is much lower than that of the liquid, so it doesn't store much heat. The key is the insulating effect of the air gap.  **(c) Animals huddle together to keep themselves warm in cold winter nights.**

**Explanation:** Animals huddle together to reduce their collective surface area exposed to the cold environment. Heat loss from a body is proportional to its surface area. By huddling, the animals minimize the total surface area exposed to the cold air, thereby reducing the rate of heat loss from their bodies to the surroundings through convection and radiation. The animals on the inside of the huddle are particularly well-insulated by the bodies of the animals surrounding them, and even the outer animals benefit from reduced exposure. This collective behavior helps conserve body heat and maintain their core body temperature in harsh conditions.

**(d) A brass tumbler feels much colder than a wooden tray on a chilly day.**

**Explanation:** The sensation of coldness or warmth depends on the rate at which heat is transferred from our body to the object (or vice-versa). Brass is a metal and is a good conductor of heat, while wood is a poor conductor of heat (an insulator). On a chilly day, both the brass tumbler and the wooden tray are at a temperature lower than our body temperature. When you touch the brass tumbler, heat rapidly flows from your warmer hand to the colder brass due to its high thermal conductivity. This rapid loss of heat from your hand creates a strong sensation of coldness. In contrast, when you touch the wooden tray, heat transfer from your hand to the wood occurs much more slowly because wood has low thermal conductivity. Therefore, your hand loses heat at a much slower rate, and the wooden tray feels less cold, or even relatively warm, compared to the brass tumbler.

Question 10

Q10. A 10 kg iron piece is dropped into 100 kg of water contained in a calorimeter of water equivalent 10 kg at 27 °C. If the final temperature is 47 °C, calculate the temperature of th

Accepted Answer

The temperature of the iron piece before dropping was 847 °C.

Question 11

Q11. A body with an initial temperature of 80 °C is cooled in a room at 25 °C. The temperature of the body drops to 70 °C in 5 minutes. What will be the temperature of the body after an

Accepted Answer

The temperature of the body after another 5 minutes (i.e., after a total of 10 minutes) will be approximately 61.3 °C.

Question 12

Q12. 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 hea

Accepted Answer

The maximum amount of ice that can melt is approximately 1.455 kg.

Question 13

Q13. In an experiment on the specific heat of a liquid, 100 g of the liquid at 100 °C is mixed with 50 g of water at 20 °C. The final temperature of the mixture is 40 °C. (Specific heat

Accepted Answer

The specific heat of the liquid is 1400 J kg⁻¹ K⁻¹.

Question 14

Q14. A refrigerator converts 100 g of water at 20 °C to ice at -10 °C in 35 minutes. Calculate the average rate of heat extraction in watts. (Specific heat of ice = 2.1 J g⁻¹ K⁻¹; Speci

Accepted Answer

The average rate of heat extraction is approximately 100.95 Watts.

Question 15

Q15. A steam engine delivers 5.4 × 10⁸ J of work per minute and consumes 3.6 × 10⁹ J of heat per minute from its boiler. Calculate the efficiency of the engine. How much heat is wasted

Accepted Answer

The efficiency of the engine is 15%. The heat wasted per minute is 3.06 × 10⁹ J.

Question 16

Q16. 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.

Accepted Answer

According to Newton's Law of Cooling, the rate of cooling of a body is directly proportional to the temperature difference between the body and its surroundings. We apply this law to both cooling intervals to find the cooling constant and then use it to determine the unknown time.

For the first interval (80 °C to 50 °C):
Average temperature of the body, T_avg1 = (80 + 50) / 2 = 65 °C
Temperature of surroundings, T_s = 20 °C
Temperature difference, ΔT1 = T_avg1 - T_s = 65 - 20 = 45 °C
Rate of cooling, (T1 - T2) / t1 = (80 - 50) / 5 = 30 / 5 = 6 °C/min

According to Newton's Law of Cooling:
(T1 - T2) / t = k (T_avg - T_s)
6 = k (45)
k = 6 / 45 = 2 / 15 min⁻¹

For the second interval (60 °C to 30 °C):
Average temperature of the body, T_avg2 = (60 + 30) / 2 = 45 °C
Temperature of surroundings, T_s = 20 °C
Temperature difference, ΔT2 = T_avg2 - T_s = 45 - 20 = 25 °C
Rate of cooling, (T3 - T4) / t2 = (60 - 30) / t2 = 30 / t2 °C/min

Using Newton's Law of Cooling again:
(T3 - T4) / t2 = k (T_avg2 - T_s)
30 / t2 = (2 / 15) * (25)
30 / t2 = 50 / 15
30 / t2 = 10 / 3
t2 = 30 * (3 / 10)
t2 = 9 minutes

Therefore, it takes 9 minutes for the body to cool from 60 °C to 30 °C.

Question 17

Q17. An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?

Accepted Answer

This problem can be solved using the First Law of Thermodynamics, which relates heat supplied, work done, and change in internal energy. Since rates are given, we use the rate form of the First Law.

Given:
Rate of heat supplied to the system, dQ/dt = 100 W = 100 J/s
Rate of work done by the system, dW/dt = 75 J/s

According to the First Law of Thermodynamics, for a system:
ΔQ = ΔU + ΔW

In terms of rates, this becomes:
dQ/dt = dU/dt + dW/dt

We need to find the rate at which the internal energy is increasing, which is dU/dt.
Rearranging the equation:
dU/dt = dQ/dt - dW/dt

Substitute the given values:
dU/dt = 100 J/s - 75 J/s
dU/dt = 25 J/s

Therefore, the internal energy of the system is increasing at a rate of 25 J/s.

Question 18

Q18. A refrigerator is to maintain eatables kept inside at 9 °C. If room temperature is 36 °C, calculate the coefficient of performance.

Accepted Answer

To calculate the coefficient of performance (COP) of a refrigerator, we need the temperatures of the cold reservoir (inside the refrigerator) and the hot reservoir (room temperature). It is crucial to convert these temperatures to the Kelvin scale before using them in the formula.

Given:
Temperature inside the refrigerator (cold reservoir), T_c = 9 °C
Room temperature (hot reservoir), T_h = 36 °C

Convert temperatures to Kelvin:
T_c = 9 + 273.15 = 282.15 K
T_h = 36 + 273.15 = 309.15 K

The coefficient of performance (COP) for a refrigerator is given by the formula:
COP = T_c / (T_h - T_c)

Substitute the Kelvin temperatures into the formula:
COP = 282.15 K / (309.15 K - 282.15 K)
COP = 282.15 K / 27 K
COP ≈ 10.45

Therefore, the coefficient of performance of the refrigerator is approximately 10.45.

Question 19

Q19. A gas is compressed adiabatically to one-third of its initial volume. If the initial pressure is 10^5 Pa and the ratio of specific heats (γ) is 1.4, calculate the final pressure.

Accepted Answer

For an adiabatic process, the relationship between pressure (P) and volume (V) is given by PV^γ = constant, where γ (gamma) is the ratio of specific heats (C_p/C_v).

Given:
Initial pressure, P₁ = 10⁵ Pa
Final volume, V₂ = V₁ / 3 (where V₁ is the initial volume)
Ratio of specific heats, γ = 1.4

We need to find the final pressure, P₂.

According to the adiabatic process equation:
P₁V₁^γ = P₂V₂^γ

Rearrange the equation to solve for P₂:
P₂ = P₁ (V₁ / V₂)^γ

Substitute the given values:
P₂ = 10⁵ Pa * (V₁ / (V₁ / 3))^1.4
P₂ = 10⁵ Pa * (3)^1.4

Now, calculate (3)^1.4:
log(3^1.4) = 1.4 * log(3)
log(3) ≈ 0.4771
1.4 * 0.4771 ≈ 0.66794
3^1.4 = antilog(0.66794) ≈ 4.655

Substitute this value back into the equation for P₂:
P₂ = 10⁵ Pa * 4.655
P₂ = 4.655 × 10⁵ Pa

Therefore, the final pressure of the gas is approximately 4.655 × 10⁵ Pa.

Question 20

Q20. A Carnot engine operates between 227 °C and 127 °C. It absorbs 6 × 10^4 J of heat from the source at 227 °C. Calculate the efficiency of the engine and the heat rejected to the sin

Accepted Answer

This problem involves a Carnot engine, which is an ideal heat engine. We need to calculate its efficiency and the heat rejected to the sink using the given temperatures and heat absorbed.

Given:
Source temperature (hot reservoir), T_h = 227 °C
Sink temperature (cold reservoir), T_c = 127 °C
Heat absorbed from the source, Q_h = 6 × 10⁴ J

First, convert temperatures to Kelvin:
T_h = 227 + 273.15 = 500.15 K ≈ 500 K
T_c = 127 + 273.15 = 400.15 K ≈ 400 K

**Part 1: Calculate the efficiency (η) of the engine**
The efficiency of a Carnot engine is given by:
η = 1 - (T_c / T_h)

Substitute the Kelvin temperatures:
η = 1 - (400 K / 500 K)
η = 1 - 0.8
η = 0.2

To express as a percentage: η = 0.2 × 100% = 20%

**Part 2: Calculate the heat rejected to the sink (Q_c)**
The efficiency can also be expressed in terms of heat:
η = (Q_h - Q_c) / Q_h = 1 - (Q_c / Q_h)

From this, we have:
Q_c / Q_h = 1 - η
Q_c = Q_h (1 - η)

Alternatively, for a Carnot engine, the ratio of heat rejected to heat absorbed is equal to the ratio of sink temperature to source temperature:
Q_c / Q_h = T_c / T_h

Using this relation:
Q_c = Q_h * (T_c / T_h)
Q_c = (6 × 10⁴ J) * (400 K / 500 K)
Q_c = (6 × 10⁴ J) * 0.8
Q_c = 4.8 × 10⁴ J

Therefore, the efficiency of the engine is 20%, and the heat rejected to the sink is 4.8 × 10⁴ J.

Question 21

Q21. A body with an initial temperature of 80 °C is cooled in a room maintained at 25 °C. The temperature of the body drops to 70 °C in 5 minutes. What will be the temperature of the bo

Accepted Answer

The temperature of the body after another 5 minutes will be approximately 61.81 °C.

Question 22

Q22. A refrigerator converts 100 g of water at 20 °C to ice at –10 °C in 35 minutes. Calculate the average rate of heat removal in watts. Given: Specific heat of ice = 2.1 J g–1 K–1, Sp

Accepted Answer

The average rate of heat removal by the refrigerator is 21 W.

Chapter 11: Thermal Properties of Matter