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 $305 ^ { \circ } \mathrm { C } \text {, }$ 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 heat lost by the metal is given by:$$Q_{lost} = m_{metal} c_{metal} (T_{initial} - T_{final})$$where:* $m_{metal}$ is the mass of the metal* $c_{metal}$ is the specific heat capacity of the metal* $T_{initial}$ is the initial temperature of the metal* $T_{final}$ is the final temperature of the metalThe heat gained by the water and the calorimeter is given by:$$Q_{gained} = (m_{water} c_{water} + m_{calorimeter} c_{calorimeter}) (T_{final} - T_{initial})$$where:* $m_{water}$ is the mass of the water* $c_{water}$ is the specific heat capacity of the water* $m_{calorimeter}$ is the mass of the calorimeter* $c_{calorimeter}$ is the specific heat capacity of the calorimeter* $T_{initial}$ is the initial temperature of the water and the calorimeter* $T_{final}$ is the final temperature of the water and the calorimeterSetting $Q_{lost}$ equal to $Q_{gained}$, we get:$$m_{metal} c_{metal} (T_{initial} - T_{final}) = (m_{water} c_{water} + m_{calorimeter} c_{calorimeter}) (T_{final} - T_{initial})$$Solving for $c_{metal}$, we get:$$c_{metal} = \frac{(m_{water} c_{water} + m_{calorimeter} c_{calorimeter}) (T_{final} - T_{initial})}{m_{metal} (T_{initial} - T_{final})}$$Plugging in the given values, we get:$$c_{metal} = \frac{(0.390 kg)(4190 J/kg ? K) + (0.090 kg)(910 J/kg ? K))(32 ^ { \circ } \mathrm { C } - 20 ^ { \circ } \mathrm { C })}{(0.160 kg)(305 ^ { \circ } \mathrm { C } - 32 ^ { \circ } \mathrm { C })}$$$$c_{metal} = 470 J/kg ? K$$

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