Question

1 Determine the quantity of heat required to convert 1kg of ice at -20
degrees Celsius to water at 100 degrees Celsius? Specific heat capacities of water and ice water are 2302 J/kgK and 4186 J/kgK respectively.

Answer 1

Answer on Question 55632, Physics, Molecular Physics | Thermodynamics

Question:

Determine the quantity of heat required to convert $1kg$ of ice at $-20{}^{\circ}\mathrm{C}$ to water at $100{}^{\circ}\mathrm{C}$? Specific heat capacities of water and ice water are $2302\frac{J}{kg\cdot K}$ and $4186\frac{J}{kg\cdot K}$ respectively.

Solution:

Let us calculate the quantity of heat that is needed to transform a $1kg$ of ice at $-20{}^{\circ}\mathrm{C}$ to water at $100{}^{\circ}\mathrm{C}$:

$$Q = Q_{1} + Q_{2} + Q_{3},$$

where $Q_{1}$ is the amount of heat required to raise the temperature of ice from $-20{}^{\circ}\mathrm{C}$ to $0{}^{\circ}\mathrm{C}$, $Q_{2}$ is the latent heat required to change the state from ice at $0{}^{\circ}\mathrm{C}$ to water at $0{}^{\circ}\mathrm{C}$ and $Q_{3}$ is the amount of heat required to raise the temperature of water from $0{}^{\circ}\mathrm{C}$ to $100{}^{\circ}\mathrm{C}$.

$$Q_{1} = m_{ice}c_{ice}\Delta t = 1kg \cdot 2302\frac{J}{kg \cdot K} \cdot (273.15K - 253.15K) = 46040J,$$

$Q_{2} = m_{ice}L_{f} = 1kg \cdot 3.33 \cdot 10^{5}\frac{J}{kg} = 333000J$ (Here, $L_{f}$ is specific latent heat of water for fusion),

$$Q_{3} = m_{water}c_{water}\Delta t = 1kg \cdot 4186\frac{J}{kg \cdot K} \cdot (373.15K - 273.15K) = 418600J.$$ $$Q = Q_{1} + Q_{2} + Q_{3} = 46040J + 333000J + 418600J = 797640J.$$

Answer:

$$Q = 797640J.$$

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