Question

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 58479, Physics, Mechanics, Relativity

Question:

Determine the quantity of heat required to convert $1\,kg$ of ice at $-20{}^{\circ}\mathrm{C}$ to water at $100{}^{\circ}\mathrm{C}$? Specific heat capacities of water and ice are $4186\,J/kg \cdot K$ and $2302\,J/kg \cdot K$ respectively. The latent heat of fusion of ice is $L_f = 3.33 \cdot 10^5\,J/kg$.

Solution:

Let's calculate the amount of heat required to convert a $1\,kg$ of ice at $-20{}^{\circ}\mathrm{C}$ to a 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 $100{}^{\circ}\mathrm{C}$.

$$Q_1 = m_{ice}c_{ice}\Delta t = 1\,kg \cdot 2302\, \frac{J}{kg{}^{\circ}\mathrm{C}} \cdot \left(0{}^{\circ}\mathrm{C} - (-20{}^{\circ}\mathrm{C})\right) = 46040\,J,$$ $$Q_2 = m_{ice}L_f = 1\,kg \cdot 3.33 \cdot 10^5\, \frac{J}{kg} = 333000\,J,$$ $$Q_3 = m_{water}c_{water}\Delta t = 1\,kg \cdot 4186\, \frac{J}{kg{}^{\circ}\mathrm{C}} \cdot (100{}^{\circ}\mathrm{C} - 0{}^{\circ}\mathrm{C}) = 418600\,J,$$ $$Q = Q_1 + Q_2 + Q_3 = 46040\,J + 333000\,J + 418600\,J = 797640\,J.$$

Answer:

$$Q = 797640\,J.$$

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