Question

Use the table of specific heat capacities to solve this problem. A bar of aluminum is heated to 90.0C and then placed into 10.0 L of water at 20.0 degrees * C . If the mixture cools to 22.4 degrees * C , what is the mass of the aluminum bar? 

Answer 1

The water will heated, while the aluminum bar will cooled. We can write the heat balance equation:

$$c_wm_w(T_{mix}-T_w)=c_{Al}m_{Al}(T_{Al}-T_{mix}).$$

From this equation we can find the mass of the aluminum bar:

$$m_{Al}=\dfrac{c_wm_w(T_{mix}-T_w)}{c_{Al}(T_{Al}-T_{mix})},$$ $$m_{Al}=\dfrac{4180\ \dfrac{J}{kg \cdot\\ ^{\circ}C}\cdot 10\ kg\cdot(22.4\ ^{\circ}C-20\ ^{\circ}C)}{900\ \dfrac{J}{kg \cdot\\ ^{\circ}C}\cdot(90\ ^{\circ}C-22.4\ ^{\circ}C)}=1.65\ kg.$$

Answer:

$m_{Al}=1.65\ kg.$