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 heat 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 capacity of the 160-g piece of metal?

Question

Quizplus · Accepted Answer

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

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