Question

determine the amount of heat energy needed to change 250g of ice at -10c into superheated stream at 130c given that the latent heat of fusion of ice is 335kj/kg the specific heat capacity of ice is 2.14kj/kg the specific heat capacity of water is 4.2kj/kg the specific heat capacity of stream is 2.01kj/kgk the latent heat of vaporisation of water is 2.26mj/kg

Answer 1

Question #68473, Engineering / Mechanical Engineering

determine the amount of heat energy needed to change 250g of ice at -10c into superheated stream at 130c given that the latent heat of fusion of ice is 335kj/kg the specific heat capacity of ice is 2.14kj/kg the specific heat capacity of water is 4.2kj/kg the specific heat capacity of stream is 2.01kj/kgk the latent heat of vaporization of water is 2.26mj/kg

Solution

Stage 1: heating the ice.

$$Q _ {1} = C m \Delta T = 2.14 \times 10 ^ {3} \times 0.25 \times 10 = 5350 \mathrm{~J}$$

Stage 2: melting ice.

$$Q _ {2} = H _ {melt} m = 335 \times 10 ^ {3} \times 0.25 = 83750 \mathrm{~J}$$

Stage 3: heating the water.

$$Q _ {3} = C m \Delta T = 4.2 \times 10 ^ {3} \times 0.25 \times 100 = 105000 \mathrm{~J}$$

Stage 4: evaporation.

$$Q _ {4} = H _ {vapor} m = 2.26 \times 10 ^ {6} \times 0.25 = 565000 \mathrm{~J}$$

Stage 3: heating the vapor.

$$Q _ {5} = C m \Delta T = 2.01 \times 10 ^ {3} \times 0.25 \times 30 = 15075 \mathrm{~J}$$

Total heat required:

$$Q = \sum Q _ {i} = 5350 + 83750 + 105000 + 565000 + 15075 = 774175 \mathrm{~J} = 774.2 \mathrm{~kJ}$$

Answer: 774.5 kJ

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