Answer
(a) aluminum
(b) $54J/s$
Work Step by Step
(a) We know that
$Q=mC_W\Delta T_W$
This can be rearranged as
$\implies C=(\frac{\Delta _W}{\Delta T})C_W$
We plug in the known values to obtain:
$C=(\frac{13}{61})(4186)=8.9\times 10^2J/Kg.K$
The specific heat shows us that the unknown material is aluminum.
(b) We can find the heating rate as
$\frac{Q}{\Delta t}=\frac{mC_w\Delta _w}{\Delta t}$
We plug in the known values to obtain:
$\frac{Q}{\Delta t}=\frac{(0.150Kg)(4186J/kg.K)(13^{\circ}C)}{2.5min(\frac{60s}
{min})}=54J/s$
