Question

5. Steam at 100°C is added to ice at 0°C. Find the temperature when the mass of steam is 10 g and the mass of ice is 50 g. The specific heat of water is 4186 J/kg°C, its latent heat of fusion is 3.33x105 J/kg and its heat of vaporization is 2.26x106 J/kg.

Answer 1

Steam converted into water by losing heat and ice absorb that heat and covert into water, after, that water will be heated to temperature T.

Mass of steam, m = 10 g

Mass of ice, M = 50 g

Heat lost = heat lost by steam = latent heat of vaporization of steam + m*(specific heat of water)*(change in temperature)

Heat lost = m*(heat of vaporization) + m*(specific heat of water)*(change in temperature)

Heat lost = 10*(2.26x106) + 10*(4186)*(373 -T) ........................................................................1

Heat gain = heat gain by ice + heat gain by water (ice + steam) to reach temperature T

Heat gain = M*heat of fusion of ice + (M + m)*(specific heat of water)*(change in temperature)

Heat gain = 50*(3.33x105) + 60*(4186)*(T - 273) .....................................................................2

Heat lost = Heat gain, by consevation of energy

So by equation 1 and 2 we get, T = 307.59 kelvin = 34.59 deg