Steam at 100 ^ o C is passed into 20g of water at 10 ^ o C . When...

Given: Specific heat of water ${S}_{w} = 1cal{g}^{- 1} ° {C}^{- 1}$, Latent heat of steam ${L}_{s} = 540cal{g}^{- 1}$

Let the mass of steam $= mg$

So heat lost in change of state from steam to water equal to sum of latent heat and heat use in raise of the temperature of water.

${Q}_{1} = m{L}_{s} + m{s}_{w} ∆ T$

${Q}_{1} = m \times 540 + m \times 1 \times ( 100 - 80 )$

${Q}_{1} = 540m + 20m = 560m$

Heat gained by water to change its temperature

${Q}_{2} = {m}_{w}{S}_{w} ∆ T$

${Q}_{2} = 20 \times 1 \times ( 80 - 10 ) = 1400$.

Now we know from the principle of calorimetry is that total heat lost by the hot body is equal to total heat gained by cold body i.e it follows law of conservation energy. Hence

Heat lost ${Q}_{1}$= Heat gained ${Q}_{2}$

$560m = 1400m = \dfrac{1400}{560} = 2 . 5g$
Therefore, total mass of water at $80 ° C$

$m = ( 20 + 2 . 5 ) g = 22 . 5g$