Question

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A reversible engine converts 1/6th of the heat input into work. When the temperature of the sink is reduced by 62 ° C, the efficiency of the engine is Doubled. What are the temperatures of source and sink?

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Answer 1

A Carnot engine converts one sixth of the heat input into work. If the sink temperature is reduced by 62o C, the efficiency gets doubled. Find the source and the sink temperature.

Maximum efficiency of an engine working between temperatures T2 and T1 is given by the fraction of the heat absorbed by an engine which can be converted into work is known as efficiency of the heat engine.
Mathematically,
Efficiency, η = (T2 – T1) / T2 = 1/6
6 T2 -6 T1 = T2
Therefore, T2 = 1.2 T1 ………………….. (1)
Where T2 is the source temperature and T1 is the sink temperature.
If the sink temperature (T1) is reduced by 62o C, the efficiency gets doubled i.e.
η= T2 – (T1 – 62) / T2 = 2 X 1/6 = 2/6
6 T2 – 6 T1 + 372 = 2 T2
6 T2 – 2 T2 – 6 T1 + 372 = 0
4 T2 – 6 T1 + 372 = 0
Substituting the value of T2 from equation (1), we will get,
4 (1.2 T1) – 6 T1 + 372 = 0
4.8 T1 – 6 T1 + 372 = 0
372 = 1.2 T1
T1 = 310 K
Therefore, T2 = 1.2 X 310 = 372 K
Hence, temperature of the source and the sink is 372 K and 310 K respectively.

Answer 2

HEY FRIEND HERE YOUR ANSWER

Efficiency η = W/Q = 1/6,

If T1 is the temperature of the source,
and T2 is the temperature of the sink,

η = (T1-T2)/T1 = 1- T2/T1

η = 1/6 = 1- T2/T1

2 η = 2/6 = (T1-(T2- 62))/T1 =1- (T2-62)/T1,
2/6 -1/6 = 1-(T2-62)/T1 – 1 - T2/T1.

1/6 = 62/T1

T1 = 372 K.

T2/T1 =1 – 1/6 = 5/6.

T2 = (5/6) •T1 = 310 K


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