A 40.0-g block of ice at - $15.00 ^ { \circ } \mathrm { C }$ is dropped into a calorimeter (of negligible heat capacity) containing water at $15.00 ^ { \circ } \mathrm { C }$ When equilibrium is reached, the final temperature is $8.00 ^ { \circ } \mathrm { C }$ How much water did the calorimeter contain initially? 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  40.0-g block of ice at - $15.00 ^ { \circ } \mathrm { C }$ is dropped into a calorimeter (of negligible heat capacity) containing water at $15.00 ^ { \circ } \mathrm { C }$ When equilibrium is reached, the final temperature is $8.00 ^ { \circ } \mathrm { C }$ How much water did the calorimeter contain initially? 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

The heat gained by the ice to reach $0 ^ { \circ } \mathrm { C }$, melt, and then warm to $8 ^ { \circ } \mathrm { C }$ must equal the heat lost by the water cooling from $15 ^ { \circ } \mathrm { C }$ to $8 ^ { \circ } \mathrm { C }$. Calculating these and solving for the mass of water gives approximately 546 g.

A 400-G Block of Ice at - 15.00∘C15.00 ^ { \circ } \mathrm { C }15.00∘C

A 400-G Block of Ice at - $15.00 ^ { \circ } \mathrm { C }$