Question

How much energy (in kilojoules) is needed to heat 4.05g of ice from -11.5∘C to 28.5∘C? The heat of fusion of water is 6.01kJ/mol, and the molar heat capacity is 36.6 J/(K⋅mol) for ice and 75.3 J/(K⋅mol) for liquid water.

Answer 1

Answer on Question #43752 - Chemistry - Inorganic Chemistry

Question:

How much energy (in kilojoules) is needed to heat 4.05g of ice from -11.5°C to 28.5°C? The heat of fusion of water is 6.01kJ/mol, and the molar heat capacity is 36.6 J/(K·mol) for ice and 75.3 J/(K·mol) for liquid water.

Solution:

Total energy needed is the sum of three items:

$$Q_{\Sigma} = Q_{1} + Q_{2} + Q_{3}$$

where $Q_{1}$ – the energy needed to heat the ice from the initial temperature ($T_{1} = -11.5^{\circ}C$) to the ice melting temperature ($T_{2} = 0.0^{\circ}C$); $Q_{2}$ – the energy needed to melt the given amount of the ice; $Q_{3}$ – the energy needed to heat the water from the melting temperature to the final temperature ($T_{3} = 28.5^{\circ}C$).

Number of moles of ice/water:

$$n = \frac{m}{M_{H_{2}O}} = \frac{4.05\,g}{18.02\,\frac{g}{mol}} = 0.225\,mol$$

The energy needed to heat the ice:

$$Q_{1} = n \cdot C_{ice} \cdot (T_{2} - T_{1}) = 0.225\,mol \cdot 36.6\,\frac{J}{mol \cdot K} \cdot (0.0 - (-11.5))K = 94.7\,J$$

The energy needed to melt the ice:

$$Q_{2} = n \cdot r = 0.225\,mol \cdot 6.01\,\frac{kJ}{mol} = 1.352\,kJ$$

The energy needed to heat the water:

$$Q_{3} = n \cdot C_{water} \cdot (T_{3} - T_{2}) = 0.225\,mol \cdot 75.3\,\frac{J}{mol \cdot K} \cdot (28.5 - 0.0)K = 482.9\,J$$

The total energy:

$$Q_{\Sigma} = Q_{1} + Q_{2} + Q_{3} = 94.7\,J + 1,352\,J + 482.9\,J = 1,929.6\,J = 1.930\,kJ$$

Answer: 1.930 kJ

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