Question

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

• A. $150K$
• B. $222K$
• C. $242K$
• D. $232K$

Answer 1

The correct option is D $232K$

Let us consider $T_1=$ temperature of the source

$T_2=$ temperature of the sink

The efficiency of the engine,

$0.25=1-\frac{T_2}{T_1}$

$\frac{T_2}{T_1}=0.75$

$T_2=0.75T_1$. . . . . . . . .(1)

When the temperature of the sink is reduced by ${58}^{0}C$, then the efficiency is double,

$2\times 0.25=1-\frac{T_2-58}{T_1}$

$\frac{T_2-58}{T_1}=0.5$

$T_2-58=0.5T_1$

$0.75T_1-58=0.5T_1$ .........(from equation 1)

$0.25T_1-58=0$

$0.25T_1=58$

$T_1=\frac{58}{0.25}$

$T_1=232K$

The correct option is D.