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Steam at $100^\circ C$ is passed into 20 g of water at $10^\circ C$. When water acquires a temperature of $80^\circ C$, the mass of water present will be: [Take specific heat of water = 1 cal $g^{-1}$ $^\circ C^{-1}$ and latent heat of steam = 540 cal $g^{-1}$]
A
24 g
B
31.5 g
C
42.5 g
D
22.5 g
Detailed Solution
Heat lost by steam = heat gained by water: $mL_v + ms_w\Delta\theta = m_Ws_W\Delta\theta_W$
$m\times 540 + m\times 1\times(100 - 80) = 20\times 1\times(80 - 10)$
$m = 2.5$ g
Total mass of water = 20 + 2.5 = 22.5 g
$m\times 540 + m\times 1\times(100 - 80) = 20\times 1\times(80 - 10)$
$m = 2.5$ g
Total mass of water = 20 + 2.5 = 22.5 g