Answer
(a) The heat capacity of the aluminum is $2.43\times 10^6~J$
(b) The heat capacity of the iron is $3.54\times 10^6~J$
Work Step by Step
(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 $3.54\times 10^6~J$
