Question

How many J of energy must be removed when 194.0 g of steam, at a temperature of 153.0 degrees Celsius, is cooled and frozen into 194.0 g of ice at 0 degrees Celsius? Take the specific heat of steam to be 2.1kJ/(kg*K).

Answer 1

Here ,

latent heat of vaporization , Lv = 2260 KJ/kg

latent heat of melting , Lf = 334 Kj/kg

specific heat of water = 4.186 KJ/kg*C

Now ,

heat removed = heat for taking steam to 100 degree C +heat to condense vapour to water + heat to take water to 0 degree C + heat to freeze water

heat removed = m *Svapour *(153 - 100) + m * Lv + m * Sw * (100 - 0) + m * Lf

heat removed = 0.194 *(2.1 *53 + 2260 + 4.186 * 100 + 334)

calculating

heat removed = 606.04 kJ

the heat removed is 606.04 kJ