How much heat is required to change 100g of water at 100°C into steam? (Latent heat of vaporization of water is $2.26 \times 〖10〗^{6} J〖kg〗^{-1}$. $(2.26\times 〖10〗^{5}J)$
Difficulty: Easy
Solution: Mass of water = m = 100 g = $\frac{100}{1000}$ = 0.1 kg
Latent heat of vaporization of water = $H_{v}$ = $2.26 \times 10^{6} Jkg^{-1}$
Heat required = $\triangle Q_{v}$= ?
$\triangle Q_{v}= mH_{v}$
$\triangle Q_{v}$= $0.1\times 2.26 \times 10^{6}$ = $0.226 \times 10^{6}$ = $\frac{226}{1000} \times 10^{6}$
= $2.26 \times 10^{-1} \times 10^{6}$ = $2.26 \times 10^{5}$J
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