Answer
(a) $24.86J/s$
(b) $19h$
Work Step by Step
(a) We can find the required rate at which heat flows into the cooler as follows:
$P=\frac{kA\Delta T}{L}$
We plug in the known values to obtain:
$P=\frac{(0.030W/m.C^{\circ})(1.5m^2)(21^{\circ}C)}{(38cm)(\frac{1m}{10^2cm})}$
$P=24.86J/s$
(b) We can find the required time as follows:
$Q=mL_f$
$Q=(5.1Kg)(33.5\times 10^4J/Kg)$
$Q=1.71\times 10^6J$
Now $t=\frac{Q}{P}$
We plug in the known values to obtain:
$t=\frac{1.71\times 10^6J}{24.86J/s}$
$t=(6.87\times 10^4s)(\frac{1h}{3600s})$
$t=19h$
