Question

What mass of steam at 100°C must be mixed with 216 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 65.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg.

Answer #1

Suppose m is the requisite mass of the steam to mix with 216 g of ice.

So, heat loss by the steam when its temperature decreases from 100 deg to 65 deg -

Q1 = m [2256 + (100 - 65) x 4.186) = m[2256 + 35 x 4.186]

and, quantity of heat gained by 216 g of ice -

Q2 = 216 * (333 + 65 x 4.186)

For the balancing condition means, heat loss = heat gain -

Equalize the two -

m[2256 + 35 x 4.186] = 216 * (333 + 65 x 4.186) = 130699.44

=> m x 2402.51 = 130699.44

=> m = 130699.44 / 2402.51 = 54.4 g.

So, mass of the requisite quantity of ice = 54.4 g.

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