A 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° C.Assume there is no external thermal energy exchange.The specific heat capacities of aluminum and water are 910 J/kg ∙ K (aluminum)and 4190 J/kg ∙ K (water).What is the specific heat of the 160-g piece of metal?

Question

Quizplus · Accepted Answer

The correct answer is A.The heat lost by the metal is equal to the heat gained by the water and the calorimeter.$$Q_{metal} = Q_{water} + Q_{calorimeter}$$$$m_{metal}c_{metal}(T_{metal} - T_{final}) = m_{water}c_{water}(T_{final} - T_{water}) + m_{calorimeter}c_{calorimeter}(T_{final} - T_{water})$$$$(0.160 kg)c_{metal}(305 ^ { \circ } \mathrm { C } - 32 ^ { \circ } \mathrm { C }) = (0.390 kg)(4190 J/kg ∙ K)(32 ^ { \circ } \mathrm { C } - 20 ^ { \circ } \mathrm { C }) + (0.090 kg)(910 J/kg ∙ K)(32 ^ { \circ } \mathrm { C } - 20 ^ { \circ } \mathrm { C })$$$$c_{metal} = 470 J/kg ∙ K$$

A 90-G Aluminum Calorimeter Contains 390 G of Water at an Equilibrium