A 44.0-g block of ice at $- 15.0 ^ { \circ } \mathrm { C }$ is dropped into a calorimeter (of neglible heat capacity)
containing 100 g of water at $5.0 ^ { \circ } \mathrm { C }$ When equilibrium is reached, how much of the ice will have
melted? 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  44.0-g block of ice at $- 15.0 ^ { \circ } \mathrm { C }$ is dropped into a calorimeter (of neglible heat capacity) 
containing  100 g  of water at $5.0 ^ { \circ } \mathrm { C }$ When equilibrium is reached, how much of the ice will have
melted? 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 lost by the ice is equal to the heat gained by the water. The heat lost 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, and $L_{f}$ is the latent heat of fusion of water.The heat gained 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, and $\Delta T_{water}$ is the change in temperature of the water.Setting $Q_{ice} = Q_{water}$, 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:$$m_{ice} = \frac{m_{water} c_{water} \Delta T_{water}}{c_{ice} \Delta T_{ice} + L_{f}}$$Substituting the given values, we get:$$m_{ice} = \frac{(0.100 \mathrm{~kg})(4186 \mathrm{~J} / \mathrm{~kg} \cdot \mathrm{~K})(5.0 \mathrm{~K})}{(2090 \mathrm{~J} / \mathrm{~kg} \cdot \mathrm{~K})(15.0 \mathrm{~K}) + (33.5 \times 10^4 \mathrm{~J} / \mathrm{~kg})}$$$$m_{ice} = 2.1 \mathrm{~g}$$Therefore, the correct answer is A.

A 440-G Block of Ice At Ice −15.0∘C- 15.0 ^ { \circ } \mathrm { C }−15.0∘C

A 440-G Block of Ice At Ice $- 15.0 ^ { \circ } \mathrm { C }$