Question

A reversible heat engine converts $\frac{1}{6}$th of heat it absorbs from source into work. When temperature of source is 600K, temperature at which heat exhausts is:

• A. 500 K
• B. 100 K
• C. 0 K
• D. 600 K

Answer 1

The correct option is A 500 K

Let the heat of source $=Q$

$\therefore$ Work done , $W=\frac{Q}{6}$

$\therefore$ Efficiency $\frac{W}{Q}=\frac{Q}{6}\times \frac{1}{Q}=\frac{1}{6}=\eta$

Now, $\eta =1-\frac{T_2}{T_1}$

$T_1=600K\therefore \frac{1}{6}=1-\frac{T_2}{600}=\frac{600-T_2}{600}\Rightarrow 100=600{-T}_2\therefore T_2=500K$

An isobaric process is a thermodynamic process in which the pressure stays constant$:\Delta P=0$. The heat transferred to the system does work, but also changes the internal energy of the system.