A 35-g block of ice at - $14 ^ { \circ } \mathrm { C }$ is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at $0 ^ { \circ } \mathrm { C }$ When the system reaches equilibrium, how much ice is left in the calorimeter? The specific heat of ice is $2090 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ that of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and the latent heat of fusion of water is $33.5 \times 10 ^ { 4 } \mathrm {~J} / \mathrm { kg }$

Question

A 35-g block of ice at - $14 ^ { \circ } \mathrm { C }$ is dropped into a calorimeter (of negligible heat capacity) containing  400 g of water at $0 ^ { \circ } \mathrm { C }$ When the system reaches equilibrium, how much ice is left in the calorimeter? The specific heat of ice is $2090 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ that of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and the latent heat of fusion of water is $33.5 \times 10 ^ { 4 } \mathrm {~J} / \mathrm { kg }$

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:* $m_{ice}$ is the mass of the ice* $c_{ice}$ is the specific heat of ice* $\Delta T_{ice}$ is the change in temperature of the ice* $L_{f}$ is the latent heat of fusion of waterThe heat lost by the water is:$$Q_{water} = m_{water} c_{water} \Delta T_{water}$$where:* $m_{water}$ is the mass of the water* $c_{water}$ is the specific heat of water* $\Delta T_{water}$ is the change in temperature of the waterSetting $Q_{ice} = Q_{water}$, we get:$$m_{ice} c_{ice} \Delta T_{ice} + m_{ice} L_{f} = m_{water} c_{water} \Delta T_{water}$$Solving for $m_{ice}$, we get:$$m_{ice} = \frac{m_{water} c_{water} \Delta T_{water}}{c_{ice} \Delta T_{ice} + L_{f}}$$Plugging in the given values, we get:$$m_{ice} = \frac{(0.400 \mathrm{~kg})(4186 \mathrm{~J} / \mathrm{~kg} \cdot \mathrm{~K})(0^\circ \mathrm{C} - (-14^\circ \mathrm{C}))}{(2090 \mathrm{~J} / \mathrm{~kg} \cdot \mathrm{~K})(0^\circ \mathrm{C} - (-14^\circ \mathrm{C})) + (33.5 	imes 10^4 \mathrm{~J} / \mathrm{~kg})}$$$$m_{ice} = 0.038 \mathrm{~kg} = 38 \mathrm{~g}$$Therefore, the correct answer is A.

A 35-G Block of Ice at - 14∘C14 ^ { \circ } \mathrm { C }14∘C

A 35-G Block of Ice at - $14 ^ { \circ } \mathrm { C }$