# 1st PUC Physics Question Bank Chapter 12 Thermodynamics

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## Karnataka 1st PUC Physics Question Bank Chapter 12 Thermodynamics

Introduction

Question 1.
What is thermodynamics?
Thermodynamics is a branch of physics which deals with concepts of heat and temperature and inter-conversion of heat and other form of energy.

Question 2.
What are thermodynamic variables?
The physical quantities such as volume, pressure, temperature and internal energy which determine the state of a system are called thermodynamic variables.

Question 3.
Give examples for thermodynamics variables.
Pressure, volume, temperature and internal energy.

Question 4.
What is thermodynamic equilibrium?
The thermodynamic variables remain in a state of equilibrium and do not change with time is called thermal equilibrium.

Thermal Equilibrium

Question .
What is thermal equilibrium?
When the temperature of two conducting bodies becomes equal, the conduction of heat in them stops. This state is known as thermal equilibrium.

Question 2.
What is the meaning of equilibrium in mechanics?
The equilibrium in mechanics means that the net external force and torque on a system are zero.

Question 3.
It is an insulating wall that does not allow flow of energy (heat).

Question 4.
What is diathermic wall?
It is a conducting wall that allows energy (heat) of flow.

Zeroth Law of Thermodynamics

Question 1.
State and explain zeroth law of thermodynamics.
If two systems are separately in thermal equilibrium with the third system, then, the two systems are also in thermal equilibrium with each other.
Explanation: Consider body ‘A’ at temperature TA and body ‘B’ at temperature TB to be in thermal equilibrium separately with the body ‘C’ at temperature TC. Then according to the law, ‘A’ and ‘B’ are in thermal equilibrium with C, TA = TC and TB = TC .This implies that TA = TB i.e., the systems A and B also in thermal equilibrium.

Heat Internal Energy and Work

Question 1.
What is heat?
Heat is a form of energy which flows from the body at a higher temperature to the body at lower temperature.

Question 2.
Give the characteristics and importance of temperature in thermodynamics?

• Temperature is a marker of the hotness of a body.
• It determines the direction of flow of heat when two bodies are placed in thermal contact.
• Heat flows from higher temperature to lower temperature.
• The heat flow stops when temperatures equalise; the two bodies are then in thermal equilibrium.

Question 3.
What is internal energy?
It is the sum of kinetic energy and potential energy of the molecules in the system. It is represented by U.
Note: Internal energy depends only on the state of the system, not on how that state was achieved.

Question 4.
Internal energy is a thermodynamic state variable. How do you account for this?
The value of internal energy depends only on the given state of the system, not on the path taken to arrive at that state. Hence internal energy is a thermodynamic state variable.

Question 5.
Which of the following statement is meaningless?
(a) A gas in a given state has certain amount of heat
(b) A gas in a given state has a certain amount of work
(c) A certain amount of heat is supplied to the system or certain amount of work done by the system.
(a) A gas in a given state has certain amount of heat.

Question 6.
What is work?
Work is the energy that is transferred from one system to another by a force moving its point of application in its own direction.
Note:
1. Heat and work are no : thermodynamic state variables but these are two different modes of altering the state of a thermodynamic system and changing its internal energy.
2. Work is done on the system (compression of gases), its internal energy increases and work is done by the system (expansion of gas), its internal energy decreases.

First Law of Thermodynamics

Question 1.
State and explain first law of thermodynamics.
Statement: When heat energy is supplied to the system, the part of this heat is utilized to increase the internal energy and the remaining heat energy is utilized to do external work by the system.
OR
The heat energy given to a system is equal to the sum of the increase in internal energy and the external work done by the system.
Explanation: Let dQ be the amount of heat supplied to the system and dU be its increased internal energy and dW be the external work done,
dQ = dU + dW
dQ = dU + PΔV

Question 2.
Discuss the applications of 1st law of Thermodynamics.
Isothermal process: A process in which the temperature of the system is kept constant throughout is called isothermal process.
Example: Slow expansion of a gas in a metallic cylinder placed in a large reservoir at constant temperature. According to first law of thermodynamics, dQ = dU + dW
In isothermal process, dU = 0 [Since temperature is constant, internal energy is also constant, change in internal energy is zero]
∴dQ = dW

Adiabatic process: A system in which no heat enters or leaves the system with the surroundings is called adiabatic process. In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero,
dQ = 0
dQ = dU + dW
∴0 = dU + dW
∴ -dU = dW

Isochoric process: The process during which volume remains constant is called isochoric constant. In isochoric process no work is done by the system, dW = 0,
dQ = dU + dW
∴dQ = dU

Isobaric process: The process during which pressure remains constant is called isobaric process.
The work done by the system at constant pressure is
dW = PdV
Therefore, dQ = dU + dW becomes
dQ = dU + PdV
W = P(V2 – V1) = µR(T2 – T1)

Cyclic process: In cyclic process, the system undergoes a series of changes and finally returns to the initial state. Since internal energy is a state function, dU = 0 for cyclic process.
So, dQ = dU + dW becomes dQ = dW i.e., the total heat absorbed equals to work done by the gas.

Specific Heat Capacity

Question 1.
Define heat capacity of a substance.
Heat capacity of a substance is defined as amount of heat required to rise the temperature unit mass of a substance by 1 kelvin.
S = $$\frac{\Delta Q}{\Delta T}$$ where ∆Q is the amount of heat supplied to the system and ∆T is the rise in temperature.

Question 2.
Define specific heat capacity(s). Give its SI unit.
The amount of heat required to raise the temperature of one kg of gas by one Kelvin is called specific heat capacity. Its SI unit is Jkg-1K-1. It is calculated by
$$s=\frac{s}{m}=\frac{1}{m} \times \frac{\Delta Q}{\Delta T}$$
where m is the mass of gas in kg.

Question 3.
Define molar heat capacity(C)? Give its SI units.
The amount of heat required to raise the temperature of one mole of gas by one kelvin is called molar heat capacity. Its SI unit is J mol-1 K-1. It is calculated by
$$C=\frac{S}{\mu}=\frac{1}{\mu} \times \frac{\Delta Q}{\Delta T}$$
where μ is the number of moles of gas.

Question 4.
What is the value of specific heat capacity of water in SI units?
4186 Jkg-1K-1

Question 5.
Define mechanical equivalent of heat.
The amount of work needed to produce one calorie of heat is called mechanical equivalent of heat.

Question 6.
Define the unit calorie.
The amount of heat required to raise the temperature of lg (1 x 10-3 kg) of water from 14.5° to 15.5°C is known as on calorie.

Thermodynamic State Variables and Equation of State

Question 1.
Name the thermodynamic state variables which describe the equilibrium state of systems.
Pressure, temperature and volume

Question 2.
Write the relation between the state variables for an ideal gas OR Write the equation of state for the ideal gas.
PV = μRT

Question 3.
What is the value of one calorie in Joules?
1 calorie = 4.186J where 4.186 is mechanical equivalent of 1 calorie of heat.

Question 4.
Define isotherm.
The pressure volume curve for a gas at fixed temperature is called isotherm.

Question 5.
Name the types of thermodynamic state variables.
There are two types of thermodynamic state variables.

1. Extensive variable
2. Intensive variable

Question 6.
What is the importance of extensive state variables? Give examples.
Extensive variables indicate the size of the system.
Examples: Internal energy, volume, mass

Question 7.
Give examples for intensive variables.
Pressure, temperature, density.

Question 8.
Divide the following into intensive and extensive variables.
(i) ΔQ (ii) ΔU (iii) PΔV (iv) T and (v) P
Extensive variables: ΔQ, ΔU, PΔV
Intensive variables: T and P

Thermodynamic Process

Question 1.
What is an ideal quasi static process?
It is an hypothetical construct in which the thermodynamic process is imagined to be slow. The variation in pressure and volume of gas takes place slowly without bringing large variation in the thermodynamic state variables of the surrounding reservoir.

Question 2.
Derive an expression CP – CV = R (Mayer’s relation).
Consider one mole of an ideal gas. Let dQ be the amount of heat absorbed by the gas. According to 1st law of thermodynamics, ΔQ = ΔU + PΔV.
If ΔQ is the amount of heat absorbed by the gas under constant volume, then by definition,
$$C_{v}=\left(\frac{\Delta Q}{\Delta T}\right)_{v}$$
Since the ideal gas depends only on temperature, and its internal energy is also depending on temperature, then
$$C_{v}=\left(\frac{\Delta Q}{\Delta T}\right)_{v} \Rightarrow C_{v}=\left(\frac{\Delta U}{\Delta T}\right)$$ ……..(1)
Let ΔQ be the amount of heat absorbed by the gas under constant pressure. Then by definition,
$$C_{p}=\left(\frac{\Delta Q}{\Delta T}\right)_{p}$$
Since the internal energy of ideal gas depends only on temperature,
$$C_{p}=\left(\frac{\Delta Q}{\Delta T}\right)_{p}$$
Since pressure is kept constant.
$$\left(\frac{\Delta Q}{\Delta T}\right)_{P}=\left(\frac{\Delta U}{\Delta T}\right)_{P}+P\left(\frac{\Delta V}{\Delta T}\right)_{P}$$
The subscript P can be dropped from the first term since V of an ideal gas depends only on T
$$C_{p}=\frac{\Delta U}{\Delta T}+P\left(\frac{\Delta V}{\Delta T}\right)$$ ……..(2)
Using (1) and (2)

Question 3.
Derive an expression for work done by the gas in a isothermal process.
Consider an ideal gas under isothermal process. For an isothermal process, PV = constant, let PV be the pressure and volume of the gas initially, then let P2V2 be the pressure and volume of the gas finally, under constant temperature, let dV be the change in volume and dW be the amount of work done by the gas.

Question 4.
Draw the P.V. curve for a isothermal process

Question 5.
Derive an expression for work done by a gas in adiabatic process.
Consider an ideal gas under adiabatic process. In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero. So work done by the gas results in decrease in its internal energy. Thus for ideal gas,
PVγ = constant
Where γ is the specific heats at constant pressure and at constant volume.
$$\gamma=\frac{C_{P}}{C_{V}}$$
If ideal gas undergoes a change in its state adiabatically from (Pi,Vj) to (P2,V2);
P1V1γ = P2V2γ
We know that dW = P x dV
PVγ = μRT

P1V1 = RT1
P2V2 = RT2

Question 6.
Draw the P-V curve for an adiabatic process.

7. Difference between Isothermal and adiabatic process.

 Isothermal Adiabatic 1. The process in which temperature remains constant. 1. The process in which temperature is not constant. 2. Exchange of heat between the surrounding and system takes place. 2. Exchange of heat between system and surrounding do not take place. 3. PV = constant 3. PVγ = constant 4. Slow process 4. Fast process 5. Example: Melting of ice 5. Example: Bursting of cycle tube

Heat Engines

Question 1.
Define heat engine.
It is a device by which a system undergoes a cyclic process and converts heat into mechanical work.

Question 2.
Discuss the parts of heat engine.

• Source (or) hot reservoir: maintained at high temperature T1 which supplies heat to working substance.
• Working substance: It goes through the series of cycle during the process of working. It absorbs Q1 amount of heat from hot reservoir and releases Q2 amount of heat to a cold reservoir which is maintained at temperature T2.
• Sink (or) cold reservoir: Maintained at a lower temperature T2 and absorbs the amount of heat energy released by working substance.

Question 3.
Draw the block diagram of part of heat engine.

Question 4.
Define efficiency of heat engine.
Ans: It is defined as the ratio of network done by the heat engine in one cycle to the amount of heat supplied to the working substance by the source.
$$\eta=\frac{W}{Q_{1}}$$
Where heat absorbed by the system in one complete cycle and W is the work done on the environment in the cycle.

Question 5.
Derive the expression for efficiency of heat engine.

Refrigerators and Heat Pumps

Question 1.
Define refrigerator.
Refrigerator is the reverse of a heat engine.

Question 2.
Draw the block diagram and write the expression for co-efficient of performance.

The coefficient of performance is given by $$\alpha=\frac{Q_{2}}{W}=\frac{Q_{2}}{Q_{1}-Q_{2}}$$
Where Q2 is the heat extracted from the cold reservoir and W is the work done on the system.

Question 3.
Give the steps involved in the working of refrigerator.
In refrigerator, the working substance involves following four steps;

• Sudden expansion of the gas from high pressure to low pressure which cools the gas and converts it into a vapour liquid mixture.
• Absorption of heat from the region to be cooled by the vapour-liquid mixture converting it vapours.
• Heating up of the vapour due to external work done on the system.
• Release of heat by the vapour to surrounding brings it to the initial state and completing the cycle.

Second Law of Thermodynamics

Question 1.
State Second law of thermodynamics.

• Kelvin – Planck statement: No process is possible whose only result is absorption of heat from reservoir and completely converted into mechanical work.
• Clausious statement: No process is possible whose sole result is transfer of heat from a colder object to a hotter object.

Reversible and Irreversible Process

Question 1.
Define thermodynamic reversible process. Give an example.
A reversible process is one which can be made to go in the reverse direction so that all changes in the direct process are exactly reversed in the opposite direction.
Example: A very slow isothermal expansion or compression of a gas is a reversible process.

Question 49.
Define thermodynamic irreversible process. Give an example.
An irreversible process is one which cannot be made to go in the reverse direction.
Example: An explosion is a irreversible process.

Carnot Engine

Question 1.
Discuss about the parts of Carnot’s engine.
Carnot engine consists of four parts

• Working substance: A cylinder containing ideal gas which is fitted with a frictionless airtight piston.
• A Source: Maintained at a higher temperature T1K which supplies heat to working substance (Q)
• Insulating stand: A device which neither accepts heat not rejects heat.
• Sink: Maintained at a lower temperature T2K which absorbs heat rejected by working substance.

Question 2.
What carnot engine?
A reversible heat engine operating between two temperatures is called carnot engine.

Question 3.
Describe the working of heat engine with an ideal gas as a working substance with the help of a Carnot cycle and hence derive the expression for efficiency of a Carnot’s engine.
Consider 1 mole of an ideal gas to be taken in a cylinder which is fitted with frictionless airtight piston.
Let P1V1T1 be its initial pressure, volume and temperature respectively.
Step 1: Let the gas be allowed to expand isothermally by keeping it on hot reservoir (source) where volume increases from V1 to V2 and temperature remains constant at T1.
Q = W1 = RT × ln($$\frac{V_{2}}{V_{1}}$$)

Step 2: The cylinder is now placed on an insulating stand and the gas is allowed to expand adiabatically so that temperature, changes from T1 to T2.
$$W_{2}=\frac{R}{\gamma-1}\left(T_{1}-T_{2}\right)$$

Step 3: The cylinder is now placed on the reservoir (sink) and the gas is allowed to compress isothermally so that volume changes from V3 to V4.
$$Q_{2}=W_{3}=-R T_{2} \ln \left(\frac{V_{3}}{V_{4}}\right)$$

Step 4: The cylinder is now kept on the insulating stand and the gas is compressed. Therefore temperature changes from T2 and T1 giving rise to an adiabatic process.
$$W_{4}=-\frac{R}{\gamma-1}\left(T_{1}-T_{2}\right)$$
Therefore total workdone

from (2) and (3)

Numerical Problems

Question 1.
A geyser heats water flowing at a rate of 3.0 litres per minute from 27°C to 77°C. If geyser operates on a gas burner, what is the rate consumption of the fuel if its the heat of combustion is 4.0 × 104 J/g.
Rate of flow of water = 3 litres / minute
∴Mass of water flowing per minute, M = 3000 g / minute
Initial temperature, T1 = 27°C
Final temperature, T2 = 77°C
Rise in temperature of the flowing water, ΔT = T2 – T1 = (77 – 27)° C = 50°C
Specific heat of water, S = 4200 J kg-1 °C-1 =4-2 Jg-1 °C-1
∴Total heat supplied by geyser, ΔQ = mSΔT
= 3000 × 4.2 × 50 = 630000 J/min = 6.3 x 105 J/min
Heat of combustion of geyser = 4 × 104 Jg-1
∴Rate of consumption of fuel, r = $$\frac{6 \cdot 0 \times 10^{5}}{4 \cdot 0 \times 10^{4}}$$ = 15.75 g / min

Question 2.
What amount of heat must be supplied to 2.0 × 10-2 kg of Nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N2 is 28; R = 8.3 J molk)
Molecular mass of N2 = 28
∴1 mole of N2 gas has a mass of 28g.
Given mass of Nitrogen = 2.0 × 10-2 kg = 20g
∴Amount of Nitrogen (in moles), n = $$\frac{20}{28}$$ moles = 0.714.
Molar specific heat capacity at constant pressure of Nitrogen, CP= $$\frac{7}{2} R$$
Rise in temperature, ΔT = 45°C
Total amount of heat to be supplied, Q = nCPΔT = 0.714 × $$\frac{7}{2}$$ x 8.3 x 45 =933.376J
Therefore, amount of heat required is 933.376 J.

Question 3.
Explain why:
(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2) / 2.
(b) The coolant in chemical or nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
(a) When two bodies at different temperatures T1 and T2 are brought in thermal contact, heat energy flows from the body at higher temperature to the body at lower temperature till thermal equilibrium is attained.
Let m be the mass of both bodies and S1 and S2 be the specific heat capacities of bodies at temperatures T1 and T2 respectively. Also, assume that T1 > T2.
Since the body at lower temperature absorbs all the energy supplied by hot body, ΔQ =-ΔQ2
∴mS1T1 = -mS2ΔT2 …….(1)
If ‘T” is the equilibrium temperature
∴mS1(T2 – T)= mS2(T2 – T) ……(2)
Re arranging equation (2), we get
T = $$\frac{S_{1} T_{1}+S_{2} T_{2}}{S_{1}+S_{2}}$$
The equilibrium temperature depends on the specific heat capacities of the two bodies and hence cannot be (T1 + T2) /2 in general.
(Note: The common temperature will be (T1 + T2)/2 only if the specific heat capacities are the same, i.e, if S1 = S2, then, T = $$\frac{\left(T_{1}+T_{2}\right)}{2}$$)

(b) The function of a coolant in a chemical or nuclear plant is to absorb as much heat as possible and prevent rise in temperature of the system. The coolant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant.
Mathematical analysis: The equilibrium temperature when two bodies come in thermal contact is given by
T = $$\frac{S_{1} T_{1}+S_{2} T_{2}}{S_{1}+S_{2}}$$ …….(1)
where S1, T1, S2, T2 represent specific heat capacity and temperature of first that of second body is respectively.
Let T2 < T1, i.e., the second body is the coolant. Re-writing (1), we have

Since the function of coolant is to keep the temperature of the system constant, the term $$\left(\frac{S_{1}}{S_{2}}\right)$$T1 in equation (2) should be as small as possible. This can be a achieved by having S2 >>S1.
Therefore, the specific head capacity of the coolant should be very high.

(C) When a car is in motion, the temperature of the tyre rises due to increases in the kinetic energy of the air molecules inside the tyre. Assuming air to be ideal gas, the equation of state is given by PV = nRT
∴ Change in the state of the system is given by
∆PV = ∆nRT …….(1)
Since there is no change in volume or composition, we can re-write equation (1) as VΔP = nRΔT
Since temperature rises during motion, ∆T is positive. Therefore, ∆P is also positive implying that pressure inside tyre increases when the car is in motion.

(d) The humidity in harbour town is generally much greater than humidity in a desert.
Since humidity is a measure of water vapour content in atmosphere and heat of water vapour is very (≈ 1.86 kJ kg-1 k-1 at 300 k) the temperature fluctuations in harbour towns are generally lower than those in desert regions. Hence, climate in harbour towns is more temperate than that of a town in a desert at the same latitude.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of cylinder are made of heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas its original volume?
Since the cylinder wall and piston are heat insulators, there is no exchange of heat. Hence, the process is “Adiabatic”.
∴The equation for the given system is P1V1ϒ= P2V2ϒ
where P1 is the initial pressure of hydrogen in cylinder, V1 is the initial volume of hydrogen in the cylinder, P2 is the final pressure of hydrogen and V2 is final volume of hydrogen in cylinder, γ is ratio of specific heat at constant to that at constant volume of hydrogen.

Thus on decreasing the volume of hydrogen to one half its original value, the pressure increases by a factor of 2.639.

Question 5.
In changing the state of gas adiabatically from equilibrium state A to another equilibrium state B, an amount of work equal to 22. 3J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the later case? (Take 1 cal = 4.19 J)
Process 1: Since the process is adiabatic, net heat supplied to the system is 0.
From the first law of thermodynamics, we have ∆Q = ∆U + ∆W … (1)
where ∆Q is the net heat supplied to the system, ∆U is the change in the internal energy of the gas and ∆W is the
net work done by the gas.
For adiabatic process, ∆Q = 0,
∴From equation (1), ∆U = -∆W
Net work done on the system is 22.3 J
i.e. ∆W = – 22.3 J
∴ U = -(- 22.3) J = 22.3 J.

Process 2: Since internal energy, U, is a state variable it does not depend on the process but only on the initial and final state of the gas. As the initial and final states for both process 1 and 2 are A and B respectively, change in internal energy is same as in Process 1.
∴ ∆U = 22.3 J …….(2)
Net heat absorbed by the system, ∆U = +9.35 cal = 9.35 × 4.19 J
∴ ∆Q = 39 176 J ……(3)
From the first law of thermodynamics,
∆Q = ∆U+ ∆W
∴∆W = ∆Q – ∆U
From equations (2) and (3),
∆W = 39.176 – 22.3 J = 16 – 876 J
.’. The net work done by the system in the second process is 16.876 J.

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is now opened. Answer the following.
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in temperature of the gas?
(d) Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface?
(a) The above process is a case of free expansion of gas. As soon as the stopcock is removed the gas expands to a total volume of twice its original value (as VA = VB).
From Boyle’s law, we have $$P \alpha \frac{1}{V}$$
Since V doubles after the stopcock is removed, P reduces to one half the original value.
At STP, pressure of the gas is 100 kPa = 1 Bar ’
∴Pressure after expansion = 50 kPa = 0.5 Bar.

(b) From first law of thermodynamics
∆Q = ∆U + ∆W
Since the process does not involves a work done by the gas such as moving piston, and no heat is exchanged,
∆Q = ∆W= 0
∴ ∆U =0
∴There is no change in internal energy of the gas.

(c) Since the internal energy of the gas is fixed for the given process, the temperature of the gas also does not change, i.e., ∆T = ∆U and if ∆U = 0, then ∆T = 0

(d) In case of free expansion of gas, the gas does not go through states of thermodynamic equilibrium before reaching the final state. Therefore, the thermodynamic parameters such pressure, volume and temperature are well defined for these intermediate states and hence, they do not lie on the R-V-T surface for the gas.

Question 7.
A steam engine delivers 5.4 × 108 J of work per minute and services 15.6 × 109 J of heat per minute from its boiler. What is the efficiency of the heat engine? How much heat is wasted per minute?
Heat supplied from boiler per minute is 3.6 × 109 J/min
Work done by the steam engine per minute = 5.4 × 108 J/min
Efficiency of steam engine,

∴ Efficiency of the steam engine is 15%
Amount of heat wasted per minute = Heat supplied to steam engine per minute – work done by the engine per minute.
= (3.6 × 109 – 5.4 × 108) J/min
= 3.06 × 109 J/min
∴ The steam engine wastes 3.06 × 109 J of energy per minute.

Question 8.
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?
Rate of heat supplied to the system is 100 W = 100 Js-1
Rate of work done by the system = 75 Js-1.
From the first law of thermodynamics, we have
∆Q = ∆U + ∆W …(1)
If the above parameters are observed for a time ‘∆t’,
$$\frac{\Delta Q}{\Delta t}=\frac{\Delta U}{\Delta t}+\frac{\Delta W}{\Delta t}$$ …….(2)
where $$\frac{\Delta Q}{\Delta t}$$ is the rate of heat supplied to the system for the given time, $$\frac{\Delta U}{\Delta t}$$ is the rate of increase in internal energy of the system for the given time and $$\frac{\Delta W}{\Delta t}$$ is the rate of work done by the system in the given time.
From equation (2), we have

Therefore the internal energy increases at rate of 25 Joules per second.

Question 9.
61. A thermodynamic system is taken from an original state D to an intermediate process shown in Figure. Its volume is then reduced to the original value from E to F by an Isobaric process. Calculate the total work done by the gas from D to E to F.

In a PV diagram,
(1) Work done for a process = Area below the curve for the given process.
∴For the given PV diagram, work done = Area of ∆DEF

∴Total work done by the gas is 450 J.

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature Is 36°C, calculate the co-efficient of performance.