JEE Main & Advanced Physics Thermodynamical Processes Sample Paper Topic Test - Thermodynamical Processes

A carnot's reversible engine converts \[\frac{1}{6}th\] of heat input into work. When the temperature of sink is reduced 62 K, the efficiency of carnot's cycle, becomes \[\frac{1}{3}.\] Calculate temperature of source and sink

A) 372K, 310K

B) 772K, 312K

C) 672K, 610K

D) None of these

Correct Answer: A

Solution :

[a] \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] where \[{{T}_{1}}\And {{T}_{2}}\] 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

\[{{T}_{1}}=372K,\,{{T}_{2}}=310\,K\]