#### Answer

$5.877 \ hours$

#### Work Step by Step

a) We know that the ice is 273 degrees Kelvin. Thus, it follows that the efficiency is:
$=100(1-\frac{273}{420})=35$%
b) The ice weighs 1,000 kilograms, so it will take $334\times 1000 = 334,000kJ$ of energy to melt it.
Using the equation for efficiency, we find:
$ Q = \frac{8.5}{.35}-8.5=15.8 \ kW$
Thus, we find:
$\Delta t = \frac{334,000}{15.8}=5.877 \ hours$