From the section Thermodynamics, you will usually get questions on heat engines or refrigerators. You will definitely remember the expression for the efficiency (η) of a Carnot engine: η = (Q1 –Q2)/Q1 = (T1–T2)/T1 where Q1 is the heat absorbed from the source at temperature T1 and T2 is the heat rejected to the sink at the lower temperature T2.
The coefficient of performance (β) of a refrigerator (heat pump) is given by
β = (Heat removed from the cold body) / (Work done by the pump)
= Q2/W = Q2/(Q1– Q2) = T2/(T1–T2)Now consider the following simple question which appeared in AIEEE 2002 question paper:
Even Carnot engine cannot give 100% efficiency because we cannot
(a) prevent radiation (b) find ideal source
(c) reach absolute zero temperature (d) eliminate frictionThe correct option is (c). The efficiency is given by the expression, η = (T1–T2)/T1. The percentage efficiency is [(T1–T2)/T1] ×100. This shows that the efficiency is 100% only if either the source temperature T1 is infinite or the sink temperature T2 is zero. Both are impossibilities.
Now see the following MCQ:
In a Carnot engine 800 J of heat is absorbed from a source at 400 K and 640 J of heat is rejected to the sink. The temperature of the sink is
(a) 320 K (b) 100 K (c) 273 K (d)250 K (e) 200 KIn a Carnot engine, Q1/T1 = Q2/T2 so that the temperature of the sink, T2 = T1Q2/Q1 = 400×640/800 = 320 K.
The following question appeared in Kerala Engineering Entrance 2000 question paper:
A heat engine undergoes a process in which its internal energy decreases by 400 J and it gives out 150 J of heat. During the process
(a) it does 250 J of work and its temperature rises
(b) it does 250 J of work and its temperature falls
(c) it does 550 J of work and its temperature rises
(d) it does 550 J of work and its temperature falls
(e) 250 J of work is done on the systemThe internal energy of the system will decrease when the system does work and/or gives off heat. Since the heat given out is 150 J and the reduction in internal energy is 400 J, the work done by the engine is 400– 150 = 250 J.
When the internal energy is reduced, the system is cooled. So, the correct option is (b).
Now, consider the following question:
The temperature inside a refrigerator is 4°C and the room temperature is 27°C. How many joules of heat will be delivered to the room for each joule of electricity consumed by the refrigerator?( Treat the refrigerator as ideal).
(a) 1 J (b) 12 J (c) 8.3 J (d) 13 J (e) 6 JThe coefficient of performance of the refrigerator, β = Q2/W = Q2/(Q1– Q2) = T2/(T1–T2) = 277/(300–277) = 12. [Note that we have converted the temperature to the Kelvin scale].
Therefore, Q2 =12 W. Heat delivered to the room is Q1 = Q2+W = 12W+W = 13W. Here W is the work done by the pump. So for each joule of work done (for each joule of electricity consumed), the quantity of heat pumped out in to the room will be 13 joule.
The coefficient of performance (β) of a refrigerator (heat pump) is given by
β = (Heat removed from the cold body) / (Work done by the pump)
= Q2/W = Q2/(Q1– Q2) = T2/(T1–T2)Now consider the following simple question which appeared in AIEEE 2002 question paper:
Even Carnot engine cannot give 100% efficiency because we cannot
(a) prevent radiation (b) find ideal source
(c) reach absolute zero temperature (d) eliminate frictionThe correct option is (c). The efficiency is given by the expression, η = (T1–T2)/T1. The percentage efficiency is [(T1–T2)/T1] ×100. This shows that the efficiency is 100% only if either the source temperature T1 is infinite or the sink temperature T2 is zero. Both are impossibilities.
Now see the following MCQ:
In a Carnot engine 800 J of heat is absorbed from a source at 400 K and 640 J of heat is rejected to the sink. The temperature of the sink is
(a) 320 K (b) 100 K (c) 273 K (d)250 K (e) 200 KIn a Carnot engine, Q1/T1 = Q2/T2 so that the temperature of the sink, T2 = T1Q2/Q1 = 400×640/800 = 320 K.
The following question appeared in Kerala Engineering Entrance 2000 question paper:
A heat engine undergoes a process in which its internal energy decreases by 400 J and it gives out 150 J of heat. During the process
(a) it does 250 J of work and its temperature rises
(b) it does 250 J of work and its temperature falls
(c) it does 550 J of work and its temperature rises
(d) it does 550 J of work and its temperature falls
(e) 250 J of work is done on the systemThe internal energy of the system will decrease when the system does work and/or gives off heat. Since the heat given out is 150 J and the reduction in internal energy is 400 J, the work done by the engine is 400– 150 = 250 J.
When the internal energy is reduced, the system is cooled. So, the correct option is (b).
Now, consider the following question:
The temperature inside a refrigerator is 4°C and the room temperature is 27°C. How many joules of heat will be delivered to the room for each joule of electricity consumed by the refrigerator?( Treat the refrigerator as ideal).
(a) 1 J (b) 12 J (c) 8.3 J (d) 13 J (e) 6 JThe coefficient of performance of the refrigerator, β = Q2/W = Q2/(Q1– Q2) = T2/(T1–T2) = 277/(300–277) = 12. [Note that we have converted the temperature to the Kelvin scale].
Therefore, Q2 =12 W. Heat delivered to the room is Q1 = Q2+W = 12W+W = 13W. Here W is the work done by the pump. So for each joule of work done (for each joule of electricity consumed), the quantity of heat pumped out in to the room will be 13 joule.
Given below is a question of the type which often finds a place in Entrance test papers:
An ideal gas is taken through a cycle of operations shown by the indicator diagram. The net work done by the gas at the end of the cycle is
(a) 6P0V0 (b) 4P0V0 (c) 15P0V0 (d) 10P0V0 (e) 3P0V0
The work done in a cyclic process indicated by a PV diagram is the area enclosed by the closed curve. The area under the slanting curve showing the expansion of the gas from volume 2V0 to volume 5V0 gives the work done by the gas. This is greater than the area under the (horizontal) curve showing the compression of the gas (from volume 5V0 to volume 2V0), which gives the work done on the gas. The vertical portion of the curve is an isochoric (volume constant) change which involves no work since the area under it is zero. The area enclosed by the closed curve gives the net work done by the gas. The triangular area enclosed is ½ ×3V0×2P0 = 3P0V0 [Option (e)].
[ Note that in problems of the above type, the work is done by the gas if the arrow showing the cycle is clockwise. If the arrow is anticlockwise, work is done on the gas. In either case, the work done is the area enclosed by the curve].
[ Note that in problems of the above type, the work is done by the gas if the arrow showing the cycle is clockwise. If the arrow is anticlockwise, work is done on the gas. In either case, the work done is the area enclosed by the curve].
Great blog sir I found it really helpful
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