# Physics Questions and Answers – Thermodynamics – Carnot Engine

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Thermodynamics – Carnot Engine”.

1. Find work done by one mole of a gas in one cycle. V1 = 1m3, V2 = 4m3, V3 = 6m3, V4 =3m3. Temperature is given in °C.

a) 303Rln(4) J
b) 303Rln(2) J
c) 30Rln(2) J
d) 30Rln(4) J

Explanation: The given T-S diagram represents a carnot cycle.
Work done is given by:
nRT1ln(V2/ V1) – nRT4ln(V3/ V4)
= 293Rln(4) – 283Rln(2)
= 303Rln(2).

2. What is the correct order of steps for a carnot cycle?

Explanation: A carnot cycle is a reversible cyclic process having the maximum possible efficiency. The steps of the cycle are: isothermal expansion of gas, adiabatic expansion, isothermal compression and then adiabatic compression. This is evident from its PV diagram:

3. What will be the value of temperature of hot reservoir in a carnot cycle, if the temperature of cold reservoir is 25°C, heat absorbed from hot reservoir is 12J and heat rejected to cold reservoir is 4J?
a) 684K
b) 333.33K
c) 894K
d) 433.33K

Explanation: For a carnot cycle: Q2/Q2 = T2/T1.
T2 = 12/4 * 298 = 894K.
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4. A carnot engine extracts 1200J of energy from a hot reservoir. What is the amount of work done by it? Temperatures of hot and cold reservoirs are 200K and 150K respectively.
a) 800J
b) 900J
c) 600J
d) 850J

Explanation: The efficiency of a carnot engine is:
1- T2/T1 = 1-150/200 = 1/4.
Efficiency is also given by: 1 – Q2/Q1 = 1 – Q2/1200 = 1/4.
Q2 = 3/4 *1200 = 900J.

5. The efficiency of a carnot engine is 50%. The temperature of the hot reservoir is kept constant. By what amount should the temperature of the cold reservoir be decreased so that efficiency becomes 60%.
a) 1K
b) 2K
c) 30K
d) 83K

Explanation: 0.5 = 1 – TC/TH.
And 0.6 = 1 – (TC + x)/TH.
From the first equation we get: 5/10 = TC/TH.
From the second equation we get: 4/10 = (TC + x)/TH.
Now, we see that the value of x must be 1K.
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