Question

Determine the amount of heat required to convert 20 gram of ice at -17°c to 20 gram water at 80°c

Answer 1

Answer on Question #65011, Physics / Molecular Physics | Thermodynamics

Question:

Determine the amount of heat required to convert 20 gram of ice at $-17{}^{\circ}\mathrm{C}$ to 20 gram water at $80{}^{\circ}\mathrm{C}$.

Solution:

There are three processes requiring heat:

1) heating the ice from $-17{}^{\circ}\mathrm{C}$ to $0{}^{\circ}\mathrm{C}$;

2) melting (= fusion) at $0{}^{\circ}\mathrm{C}$;

3) heating the water from $0{}^{\circ}\mathrm{C}$ to $80{}^{\circ}\mathrm{C}$.

So, we need to use these constants:

1) heat capacity of ice $C_i = 2.05 \frac{J}{g \cdot K}$

2) heat of fusion $H = 334 \frac{J}{g}$

3) heat capacity of ice $C_w = 4.18 \frac{J}{g \cdot K}$

Total amount of heat may be calculated this way:

$$\begin{array}{l} H_{tot} = 20 g \cdot C_i \cdot \left(0{}^{\circ}\mathrm{C} - (-17{}^{\circ}\mathrm{C})\right) + 20 g \cdot H + 20 g \cdot C_w (80{}^{\circ}\mathrm{C} - 0{}^{\circ}\mathrm{C}) = \\ = 20 \cdot 2.05 \cdot 17 + 20 \cdot 334 + 20 \cdot 4.18 \cdot 80 = 14065 \mathrm{~J} \end{array}$$

Answer:

14065 J

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