A 44.0-g block of ice at -15.0°C is dropped into a calorimeter (of negligible heat capacity)containing $100 \mathrm {~g}$ of water at 5.0°C.When equilibrium is reached,how much of the ice will have melted? The specific heat of ice is 2090 J/kg ∙ K,that of water is 4186 J/kg ∙ K,and the latent heat of fusion of water is 33.5 × 10⁴ J/kg.

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Quizplus · Accepted Answer

Explanation:The heat gained by the ice is equal to the heat lost by the water. The heat gained by the ice is:$$Q_{ice} = m_{ice}c_{ice}\Delta T_{ice} + m_{ice}L_f$$where:$$m_{ice} = 44.0 \mathrm{~g} = 0.0440 \mathrm{~kg}$$$$c_{ice} = 2090 \mathrm{~J/kg \cdot K}$$$$\Delta T_{ice} = 15.0 \mathrm{~°C} - 0 \mathrm{~°C} = 15.0 \mathrm{~°C}$$$$L_f = 33.5 \times 10^4 \mathrm{~J/kg}$$Substituting these values into the equation, we get:$$Q_{ice} = (0.0440 \mathrm{~kg})(2090 \mathrm{~J/kg \cdot K})(15.0 \mathrm{~°C}) + (0.0440 \mathrm{~kg})(33.5 \times 10^4 \mathrm{~J/kg})$$$$Q_{ice} = 13.5 \mathrm{~kJ}$$The heat lost by the water is:$$Q_{water} = m_{water}c_{water}\Delta T_{water}$$where:$$m_{water} = 100 \mathrm{~g} = 0.100 \mathrm{~kg}$$$$c_{water} = 4186 \mathrm{~J/kg \cdot K}$$$$\Delta T_{water} = 5.0 \mathrm{~°C} - 0 \mathrm{~°C} = 5.0 \mathrm{~°C}$$Substituting these values into the equation, we get:$$Q_{water} = (0.100 \mathrm{~kg})(4186 \mathrm{~J/kg \cdot K})(5.0 \mathrm{~°C})$$$$Q_{water} = 2.09 \mathrm{~kJ}$$Since the heat gained by the ice is equal to the heat lost by the water, we have:$$Q_{ice} = Q_{water}$$$$13.5 \mathrm{~kJ} = 2.09 \mathrm{~kJ}$$Solving for $m_{ice}$, we get:$$m_{ice} = \frac{Q_{water}}{L_f} = \frac{2.09 \mathrm{~kJ}}{33.5 \times 10^4 \mathrm{~J/kg}} = 0.00624 \mathrm{~kg} = 6.24 \mathrm{~g}$$Therefore, the amount of ice that will have melted is 6.24 g.

A 440-G Block of Ice at -15 100 g100 \mathrm {~g}100 g Of Water at 5

A 440-G Block of Ice at -15 $100 \mathrm {~g}$ Of Water at 5