Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be [Take specific heat of water = 1 cal g-1oC and latent heat of steam = 540 cal g-1 ]

Here, $Specific heat of water, s_{w} = 1 cal {g^{- 1}}^{\circ} C^{- 1}$

$Latent heat of steam, L_{s} = 540 cal g^{- 1}$

$\begin{aligned} Heat lost by m g of steam at 100^{\circ} C to change into water at 80^{\circ} C is \end{aligned}$

$\begin{aligned} Q_{1} = m L_{s} + m s_{w} Δ T_{w} \\ = m \times 540 + m \times 1 \times (100 - 80) = 540 m + 20 m = 560 m \end{aligned}$

$Heat gained by 20 g of water to change its temperature$ $from 10^{\circ} C to 80^{\circ} C is$

$Q_{2} = m_{w} s_{w} Δ T_{w} = 20 \times 1 \times (80 - 10) = 1400$

$According to principle of calorimetry$

$\begin{array}{l} Q_{1} = Q_{2} \\ ∴ 560 m = 1400 or m = 2.5 g \end{array}$

Total mass of water present

$= (20 + m) g = (20 + 2.5) g = 22.5 g$

Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be [Take specific heat of water = 1 cal $g^{- 1 o} C$ and latent heat of steam = 540 cal $g^{- 1}$ ]