## College Physics (4th Edition)

(a) The heat capacity of the aluminum is $2.43\times 10^6~J$ (b) The heat capacity of the iron is $3.54\times 10^6~J$
(a) We can find the mass of the aluminum: $m = (2700~kg/m^3)(1.00~m^3) = 2700~kg$ The heat capacity is the amount of heat required to raise the temperature by $1~ C^{\circ}$. We can find the heat capacity of the aluminum: $Q = m~c~\Delta T$ $Q = (2700~kg)(900~J/kg~C^{\circ})(1~C^{\circ})$ $Q = 2.43\times 10^6~J$ The heat capacity of the aluminum is $2.43\times 10^6~J$ (b) We can find the mass of the iron: $m = (7860~kg/m^3)(1.00~m^3) = 7860~kg$ The heat capacity is the amount of heat required to raise the temperature by $1~ C^{\circ}$. We can find the heat capacity of the aluminum: $Q = m~c~\Delta T$ $Q = (7860~kg)(450~J/kg~C^{\circ})(1~C^{\circ})$ $Q = 3.54\times 10^6~J$ The heat capacity of the iron is $3.54\times 10^6~J$