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

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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]

1. 24 g
2. 31.5 g
3. 42.5 g
4. 22.5 g

Correct Option: D

According to the principle of calorimetry.
Heat lost = Heat gained
mL_v + ms_w∆θ = m_ws_wΔ θ
⇒ m × 540 + m × 1 × (100 – 80)
= 20 × 1 × (80 – 10)
⇒ m = 2.5 g
Therefore total mass of water at 80°C
= (20 + 2.5) g = 22.5 g