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Question: An engine has an efficiency of 0.25 when temperature of sink is reduced by 58°C, if its efficiency i...

An engine has an efficiency of 0.25 when temperature of sink is reduced by 58°C, if its efficiency is doubled, then the temperature of the source is

A

150°C

B

222°C

C

242°C

D

232°C

Answer

232°C

Explanation

Solution

: Here, η1=1−T2T1\eta _ { 1 } = 1 - \frac { T _ { 2 } } { T _ { 1 } } Or 0.25=1−T2 T1⇒14=1−T2 T10.25 = 1 - \frac { \mathrm { T } _ { 2 } } { \mathrm {~T} _ { 1 } } \Rightarrow \frac { 1 } { 4 } = 1 - \frac { \mathrm { T } _ { 2 } } { \mathrm {~T} _ { 1 } }

T2 T1=1−14=34\frac { \mathrm { T } _ { 2 } } { \mathrm {~T} _ { 1 } } = 1 - \frac { 1 } { 4 } = \frac { 3 } { 4 }

According to question

η2=2η1\eta _ { 2 } = 2 \eta _ { 1 } and T2=T2−58∘C\mathrm { T } _ { 2 } = \mathrm { T } _ { 2 } - 58 ^ { \circ } \mathrm { C }

∴2×14=1−(T2−58∘C)T1⇒1−12=T2−58∘CT1\therefore 2 \times \frac { 1 } { 4 } = 1 - \frac { \left( \mathrm { T } _ { 2 } - 58 ^ { \circ } \mathrm { C } \right) } { \mathrm { T } _ { 1 } } \Rightarrow 1 - \frac { 1 } { 2 } = \frac { \mathrm { T } _ { 2 } - 58 ^ { \circ } \mathrm { C } } { \mathrm { T } _ { 1 } }