Question

An engine is working. It takes $100$ calories of heat from source and leaves $80$ calories of heat to sink. If the temperature of source is $127^{\circ }C$, then temperature of sink is

• A. $147^{\circ }C$
• B. $47^{\circ }C$
• C. $100^{\circ }C$
• D. $47K$

Answer 1

Answer: (B)

Heat taken from source $Q_1=100$ cal
Heat left to sink $Q_2=80$ cal
$\therefore$ efficiency of the engine $\eta =1-\frac{Q_2}{Q_1}$
$=1-\frac{80}{100}=20\%$
Temperature of the source $T_1=127ºC=400K$
Temperature of the sink $T_2=?$
We know that
$\eta =1-\frac{T_2}{T_1}$
$\Rightarrow 0.2=1-\frac{T_2}{400}$
$\Rightarrow \frac{T_2}{400}=1-0.2=0.8$
$\Rightarrow T_2=320K=47^{\circ }C$