Efficiently Manufacture 3.0kg of Ice Cubes: Calculating Minimum Work

[CartoonKid]

Calculate the minimum work required to manufacture 3.0kg of ice cubes from water initially at 0C. Assume that the surroundings are at 30C. The latent heat of fusion of water at 0C is 338.7kJ/kg.

My working:
e=1-\frac{273}{303}
then
e=\frac{work}{input}
input=(3.0)(338.7E3)
work=100.6kJ

But I saw the other answer from someone that it is 117.1kJ. Which one is the correct answer? Is my answer correct? If not, please rectify for me. Thanks.

 

[ColdRifle]

First of all efficiency=1-T2/T1*100%=1-273/303*100%=9.9%
Then efficiency=usefuloutput/Inputbythemachine=(3*338700)/input
So
work needed as I calculated is 102.64KJ AND you are both wrong (if I am not wrong...)
Thats because what you have to calculate is the input the machine needs since it has the given efficiency to make it out to create ice cubes or whatever...

 

[CartoonKid]

Now I have got the way of how the other answer was being worked out.
COP=\frac{Q_{cold}}{Work}
COP=\frac{T_{cold}}{T_{hot}-T_{cold}}
After sub in all the values,
Work=\frac{(3.0)(338.7E3)}{9.1}
and the answer appears to be 111.7kJ (sorry, typo error in the first post)
ColdRifle, I don't understand why you sub in (3.0)(338.7E3) as your usefuloutput. Isn't that eff=work/input for carnot cycle?