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 and latent heat of steam = 540 cal ] [2014]

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Solution

Q.
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 $g^{- 1}$ $° C^{- 1}$ and latent heat of steam = 540 cal $g^{- 1}$ ] [2014]

1 24 g

2 31.5 g

3 42.5 g

4 22.5 g

Ans.
(4)

Here,

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

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

Heat lost by $m$ g of steam at $100^{o} C$ to change into water at $80 °$ C is

$Q_{1} = m L_{s} + m s_{w} Δ T_{w}$

$= m \times 540 + m \times 1 \times (100 - 80)$

$= 540 m + 20 m = 560 m$

Heat gained by 20 g of water to change its temperature from $10 °$ C to $80 °$ C is

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

According to the principle of calorimetry, $Q_{1} = Q_{2}$

$∴$ $560 m = 1400 \Rightarrow m = 2.5$ $g$

Total mass of water present

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

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