10125 J of heat energy boils off 4.5 g of water at 100°C to steam at 100°C. Find the specific latent heat of steam.

Given,

Mass (m) = 4.5 g = $\dfrac{4.5}{1000}$ = 4.5 x 10^-3 kg

Heat energy given out (Q) = 10,125 J

Specific latent heat of steam = ?

From relation Q = m x L

Substituting the values in the formula we get,

$10125 = 4.5 \times 10^{-3} \times L \\[0.5em] \Rightarrow L = \dfrac{10125}{4.5 \times 10^{-3}} \\[0.5em] \Rightarrow L = 2.25 \times 10^{6} \text{ J kg}^{-1} \\[0.5em]$

Hence, the specific latent heat of steam = 2.25 x 10⁶ J kg^-1