Question

Steam at $100^\circ C$ is passed into $22g$ of water at $20^\circ C$ . The mass of water that will be present when the water acquires a temperature of $90^\circ C$ (Latent heat of steam is $540calg^{- 1}$ ) is

• A. $24.8gm$
• B. $24gm$
• C. $36.6gm$
• D. $30gm$

Answer 1

Answer: (A)

Heat loss $=$ Heat gain
$mL_{V}+mS_{W}\left(100 - 90\right)=22S_{w}\left(90 - 20\right)$
$m\left[\right.540+10\left]\right.=22\times 1\times 70$
$m=\frac{22 \times 70}{550}=2.8gm$
Mass of water $=22+2.8=24.8gm$