Question

A sample of $0.1g$ of water at ${100}^{o}C$ and normal pressure $(1.013\times {10}^{5}N{m}^{-2})$ requires $54$ cal of heat energy of convert to steam at ${100}^{o}C$. If the volume of the steam produced is $167.1cc$, the change in internal energy of the sample, is?

• A. $42.2J$
• B. $104.3J$
• C. $84.5J$
• D. $208.7J$

Answer 1

Answer: (D)

$208.7J$

From first law of thermodynamics,

$\delta Q=dU+\delta W$

$54\times 4.184=dU+PdV$

$54\times 4.184=dU+1.013\times {10}^5\times 167.1\times {10}^{-6}0$

On solving, we will get $dU=208.7J$