11A 90-g aluminum calorimeter contains 390 g of water at an equilibrium temperature of $20 ^ { \circ } \mathrm { C }$ A 160-g piece of metal, initially at $305 ^ { \circ } \mathrm { C }$ is added to the calorimeter. The final temperature at equilibrium is $32 ^ { \circ } \mathrm { C }$ Assume there is no external heat exchange. The specific heat capacities of aluminum and water are 910 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K }$ (aluminum) and 4190 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K }$ (water). What is the specific heat capacity of the 160-\mathrm{g} piece of metal?

Question

11A  90-g aluminum calorimeter contains  390 g  of water at an equilibrium temperature of $20 ^ { \circ } \mathrm { C }$ A 160-g piece of metal, initially at $305 ^ { \circ } \mathrm { C }$ is added to the calorimeter. The final temperature at equilibrium is $32 ^ { \circ } \mathrm { C }$ Assume there is no external heat exchange. The specific heat capacities of aluminum and water are 910 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K }$ (aluminum) and 4190 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K }$   (water). What is the specific heat capacity of the  160-\mathrm{g}  piece of metal?

Quizplus · Accepted Answer

The heat gained by the water and calorimeter equals the heat lost by the metal. Using $Q = mc\Delta T$ for each, and solving for the metal's specific heat capacity $c$, we find $c = 470 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$.

11A 90-G Aluminum Calorimeter Contains 390 G 20∘C20 ^ { \circ } \mathrm { C }20∘C

11A 90-G Aluminum Calorimeter Contains 390 G $20 ^ { \circ } \mathrm { C }$