Question

Calculate the total heat required to completely turn 20g of ice at -20°c to steam​

Answer 1

Answer:

The total heat required to completely turn 20 g of ice at -20°C to steam, is 65880 J

Step-by-step explanation:

The given parameters are;

The mass of the ice = 20 g

The initial temperature of the ice = -20°C

The specific heat capacity of the ice, ΔH(fusion) = 2.03 J/(g·°C)

The latent heat of fusion of ice = 334 J/g

The specific heat capacity of the water = 4.184 J/(g·°C)

The latent heat of evaporation of water = 2501 J/g

The heat required to raise the temperature of the ice to from -20°C to 0°C, Q₁, is given as follows;

Q₁ = 20 × 2.03 × (0 - (-20)) = 812

Q₁ = 812 J

The latent heat required to melt the 20 g of ice at 0°C is given as follows;

Q₂ = ΔH(fusion) × mass, m = 334 × 20 = 6680

Q₂ = 6680 J

The heat, Q₃, required to raise the temperature of the ice to 100°C (boiling point temperature) is given as follows;

Q₃ = 20 × 4.184 × (100 - 0) = 8368

Q₃ = 8368 J

The heat, Q₄, required to evaporate the 20 g of water at 100°C is given as follows

Q₄ = 2501 × 20 = 50020

Q₄ =  50020 J

The total heat, $Q_{total}$ required to completely turn 20 g of ice at -20°C to steam, is given as follows;

$Q_{total}$ = Q₁ + Q₂ + Q₃ + Q₄ =  812 + 6680 + 8368 + 50020 = 65880

$Q_{total}$ = 65880 J.