Answer
$\Delta S = 1.18 \times 10^3 J/K$
Work Step by Step
To find the entropy when ice is warmed from $-20^oC$ to $0^oC$ , use the following equation
$\Delta S_{1} = mC_i ln (\frac{T_f}{T_i}) $
$\Delta S_{1} = (0.60kg)(2200J/kg.K) ln (\frac{273K}{253K}) $
$\Delta S_{1} = 101.34 J/K$
Now find the entropy when the ice melts at $0^oC$
$\Delta S_{2} = \frac{mL_f}{T}$
$\Delta S_{2} = \frac{(0.60 kg )(333000 J/kg)}{273 K} $
$\Delta S_{2} = 731. 87 J/K$
Now calculate the warming of water from $0^oC$ to $40^oC$. Change it to Kelvin first.
$ \Delta S_{3} = mC_w ln (\frac{T_f}{T_i}) $
$ \Delta S_{3} = (0.60 kg) (4190 J/kg.K) ln (\frac{313K}{273K}) $
$ \Delta S_{3} = 343.74 J/K$
So the total entropy change is
$\Delta S = \Delta S_1 + \Delta S_2 + \Delta S $
$\Delta S = 101.34 J/K + 731. 87 J/K + 343.74 J/K $
$\Delta S = 1176.95 J/K = 1.18 \times 10^3 J/K$
The answer is rounded off to the nearest two decimal places.
