Themo Problem

[eku_girl83]

Here's my problem:
For each kW of power input to an icemaker at steady state, determine the maximum rate that ice can be produced in kg/h, from liquid water at 0 C. Assume that 333 kJ/kg of energy must be removed by heat transfer to freeze water at 0 C, and that the surroundings are at 20 C.

I know that W = 1 kw, Tc = 0C=273 K, Th = 20C=293 K.
Since the device operates at steady state, W = Qh - Qc.
I also know that W/Qh is less than or equal to (1- Tc/Th)

Could someone give me a hint on where to start with this one? What is the 333 kJ/kg?

Thanks!

[CartoonKid]

The answer is 14.76kgh^{-1}, is it?

My working is as follow:
Q_c=mL_f=m(333\times10^3)
\frac{Q_c}{Q_h}=\frac{T_c}{T_h}
\Rightarrow Q_h=m(357.4\times10^3)
W=Q_h-Q_c=m(24.4\times10^3)
1\times10^3=\frac{m(24.4\times10^3)}{t}
\frac{m}{t}=40.98\times10^{-3}kgs^{-1}=147.54kgh^{-1}