Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$5.25\times 10^4~J$ of heat energy must be added to the block of aluminum.
We can find the mass of the block as: $m = \rho~V$ $m = (2700~kg/m^3)(0.060~m)^3$ $m = 0.5832~kg$ The heat energy $Q$ required to raise the temperature of a substance is: $Q = m~c~\Delta T$ We then find the heat energy required to raise the temperature of the aluminum block: $Q = m~c~\Delta T$ $Q = (0.5832~kg)(900~J/kg~C^{\circ})(100~C^{\circ})$ $Q = 5.25\times 10^4~J$ $5.25\times 10^4~J$ of heat energy must be added to the block of aluminum.