Question

$540$ calories of heat convert $1\text{cc}$ of water at ${100}^{\circ }\text{C}$ to $1670\text{cc}$ steam at ${100}^{\circ }\text{C}$ at a pressure of $1\text{atm}$. Then the work done against the atmospheric pressure is nearly (Take atmospheric pressure, $P=1.01\times {10}^6{\text{dyne/cm}}^2$)

• A. $40\text{cal}$
• B. $30\text{cal}$
• C. $16.7\text{cal}$
• D. $10\text{cal}$

Answer 1

The correct option is A $40\text{cal}$

Given,
Amount of heat supplied $=540\text{cal}$
Change in volume $\left(\Delta V\right)=1670\text{cc}$
We know that,
Atmospheric pressure $P=1.01\times {10}^6{\text{dyne/cm}}^2$
Work done against atmospheric pressure,
$W=P\Delta V=\frac{1.01\times {10}^6\times 1670}{4.2\times {10}^7}=40\text{cal}$