Question

How much heat is needed to turn the 1 kg of ice at -4°C to turn it into vapor at 100°C?

Answer 1

Answer on Question #79452 - Physics - Molecular Physics, Thermodynamics

How much heat is needed to turn the 1 kg of ice at -4°C into vapor at 100°C?

Solution.

The necessary constants:

Specific heat capacity of ice $c_{\text{ice}} = 2.09 \, \text{kJ/(kg °C)}$

(https://www.engineeringtoolbox.com/specific-heat-solids-d_154.html)

Specific heat capacity of water $c_{\text{water}} = 4.19 \, \text{kJ/(kg °C)}$

(https://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html)

Latent heat of fusion of water $\lambda_{\text{water}} = 334 \, \text{kJ/kg} = 334 \, \text{kJ/kg}$

Latent heat of vaporization of water $L_{\text{water}} = 2264 \, \text{kJ/kg}$

(https://en.wikipedia.org/wiki/Latent_heat)

The process consists of four stages:

1. Heating of ice to its melting point (0°C).

2. Ice melting (the temperature remains constant until all of the ice turns to water).

3. Heating of water to its boiling temperature (100°C).

4. Vaporization of water at a constant temperature.

The energy required to heat the ice of mass $m = 1 \, \text{kg}$ from $t_1 = -4{}^\circ \text{C}$ to $t_2 = 0{}^\circ \text{C}$:

$$Q_1 = c_{\text{ice}} m(t_2 - t_1) = 2.09 \frac{kJ}{kg \, {}^\circ \text{C}} \times 1 \, \text{kg} \times 4{}^\circ \text{C} = 8.36 \, \text{kJ}$$

The amount of heat needed to melt the ice:

$$Q_2 = \lambda_{\text{water}} m = 334 \frac{kJ}{kg} \times 1 \, \text{kg} = 334 \, \text{kJ}$$

The energy that is necessary to heat water from $t_2 = 0{}^\circ \text{C}$ to $t_3 = 100{}^\circ \text{C}$:

$$Q_3 = c_{\text{water}} m(t_3 - t_2) = 4.19 \frac{kJ}{kg \, {}^\circ \text{C}} \times 1 \, \text{kg} \times 100{}^\circ \text{C} = 419 \, \text{kJ}$$

The heat required to turn all the water into vapor:

$$Q_4 = L_{\text{water}} m = 2264 \frac{kJ}{kg} \times 1 \, \text{kg} = 2264 \, \text{kJ}$$

The total heat needed to turn 1 kg of ice at -4°C into vapor at 100°C:

$$Q = Q_1 + Q_2 + Q_3 + Q_4 = (8.36 + 334 + 419 + 2264) \, \text{kJ} = 3025.36 \, \text{kJ} = 3025.36 \times 10^3 \, \text{J}$$

Answer: $Q = 3025.36 \, \text{kJ}$.

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