Answer
(a) $1.30\times 10^3J$, it is 1.29 times more than the specific heat.
(b) $4.19\times 10^6J$, it is 1000 times more than the specific heat.
Work Step by Step
(a) We know that
$Q=mC\Delta T$
$\implies Q=\rho V C\Delta T$ (where $m=\rho V$)
We plug in the known values to obtain:
$Q=(1.29Kg/m^3)(1.00m^3)[1004J/(Kg.K)](1.00^{\circ}C)$
$Q=1.30\times 10^3J$
Thus, it is 1.29 times more than the specific heat.
(b) We know that
$Q=mC\Delta T$
$\implies Q=\rho V C\Delta T$ (where $m=\rho V$)
We plug in the known values to obtain:
$Q=(1000Kg/m^3)(1.00m^3)[4186J/(Kg.K)](1.00^{\circ}C)$
$Q=4.19\times 10^6J$
Thus, it is 1000 times more than the specific heat.
