Answer
$t =1.20h$
Work Step by Step
To find the time taken, we first find the heat energy in the system
$Q = mc\Delta T$
$Q =(12.0 kg)(4.19×10^3 J/kg⋅^oC)(5.0°C − 31°C) $
$Q = -1.31 \times 10^6 J$ And the magnitude is $|Q_C|$
Coefficient of performance, $K = \frac{𝑄_C}{𝑊}$ and $W = Pt$
$K = \frac{|Q_C|}{Pt}$
$t = \frac{|Q_C|}{PK}$
$t = \frac{ 1.31 \times 10^6 J}{(135 W)(2.25)}$
$t = 4313 s = 71.88 min = 1.20h$
