SATHEE: Thermodynamics Question 307

Thermodynamics Question 307

Question: A carnots reversible engine converts $\frac{1}{6} t h$ of heat input into work. When the temperature of sink is reduced 62 K, the efficiency of carnotss cycle, becomes $\frac{1}{3} .$ Calculate temperature of source and sink

Options:

A) 372K, 310K

B) 772K, 312K

C) 672K, 610K

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $η = 1 - \frac{T_{2}}{T_{1}}$

where $T_{1} & T_{2}$

are source & sink temperatures respectively. $\frac{1}{6} = 1 - \frac{T_{2}}{T_{1}}$ ..(1)

Modified case,

$\frac{1}{3} = 1 - \frac{T_{2} - 62}{T_{1}}$ …(2)

Solving (1) and (2), we get

$T_{1} = 372 K, T_{2} = 310 K$

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Thermodynamics Question 307

Question: A carnots reversible engine converts 16th of heat input into work. When the temperature of sink is reduced 62 K, the efficiency of carnotss cycle, becomes 13. Calculate temperature of source and sink

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Question: A carnots reversible engine converts $\frac{1}{6} t h$ of heat input into work. When the temperature of sink is reduced 62 K, the efficiency of carnotss cycle, becomes $\frac{1}{3} .$ Calculate temperature of source and sink