Question

Steam at $100°\text{C}$ is passed into $20\text{g}$ of water at $10°\text{C}$. When water acquires a temperature of $80°\text{C}$, the mass of water present will be: [Take specific heat of water $=1{\text{cal g}}^{-1}{\text{C}}^{-1}$ and latent heat of steam $=540{\text{cal g}}^{-1}$]

• A. $24\text{g}$
• B. $31.5\text{g}$
• C. $42.5\text{g}$
• D. $22.5\text{g}$

Answer 1

The correct option is D $22.5\text{g}$

According to the principle of calorimetry
Heat lost = Heat gained
$m{L}_{\text{vapourisation}}+m{s}_w\Delta T=m_w{s}_w\Delta T$
$m\times 540+m\times 1\times (100-80)=20\times 1\times (80-10)$
$m=2.5\text{g}$
Total mass of water $=(20+2.5)\text{g}=22.5\text{g}$