Question

A sample of $0.1\text{g}$ of water at ${100}^{\circ }\text{C}$ and normal pressure $(1.03\times {10}^5{\text{Nm}}^{-2})$ requires $54\text{cal}$ of heat energy to convert to steam at ${100}^{\circ }\text{C}$. If the volume of the steam produced is $167.1\text{cc}$, the change in internal energy of the sample is

• A. $+104.4\text{J}$
• B. $+208.8\text{J}$
• C. $-84.5\text{J}$
• D. $-242.6\text{J}$

Answer 1

The correct option is B $+208.8\text{J}$

Given,
Heat required by water $\left(\Delta Q\right)=+54\text{cal}$
$\Rightarrow \Delta Q=54\times 4.18=+225.72\text{J}$
It is given that pressure is constant and is equal to atmopheric pressure, $P=1.013\times {10}^5\text{Pa}$.
Volume of water is $\frac{m}{\rho }=\frac{0.1\text{g}}{1\text{g/cc}}=0.1\times {10}^{-6}{\text{m}}^3$
Change in volume $\left(\Delta V\right)=(167.1\times {10}^{-6}-0.1\times {10}^{-6})=+167\times {10}^{-6}{\text{m}}^3$
Now, as per first law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q=\Delta U+P\Delta V$
$\Rightarrow 225.72=\Delta U+1.013\times {10}^5(167\times {10}^{-6})$
$\Rightarrow \Delta U=+208.8\text{J}$
Thus, option (b) is the correct answer.