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If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is
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- 270ºC
- 230ºC
- 100ºC
- 50ºC
Correct Option: C
Heat required by ice at 0°C to reach a temperature of 100°C = mL + mc∆θ
= 1 × 80 + 1 × 1 × (100 – 0) = 180 cal
Heat available with 1 g steam to condense into 1 g of water at 100°C = 536 cal.
Obviously the whole steam will not be condensed and ice will attain a temperature of 100°C; so the temperature of mixture = 100 °C
