Question

A heat engine rejects 600 cal to the sink at ${27}^{o}C$. Amount of work done by the engine will be
(Temperature of source is ${227}^{o}C$ &$J$ =$4.2J/cal)$

• A. 1680 J
• B. 840 J
• C. 2520 J
• D. None of these

Answer 1

The correct option is A 1680 J

We know,

$\frac{Q_1}{Q_2}=\frac{T_1}{T_2}$

$\Rightarrow Q_2=600$ cal

$\Rightarrow T_1=227+273=500$K

$\Rightarrow T_2=27+273=300$K

Putting these values

$Q_1=Q_2\times \frac{T_1}{T_2}=600\times \frac{500}{300}=1000$cal

Also, we know,

Efficiency

$\Rightarrow n=\frac{W}{Q_1}=1-\frac{T_1}{T_2}$

$W=Q_1[1-\frac{T_1}{T_2}]=1000[1-\frac{300}{500}]$

$W-400$cal

Thus, work done will be $W=400\times 4.2J=1680J$.