# Homework Help: Thermodynamics-reversible carnot engine

1. Mar 2, 2015

### anisha

1. The problem statement, all variables and given/known data
a refrigerator is used to convert 1 kg of liquid water at 30 degree celsius to ice at -18 degree celsius.the refrigerator rejects heat to surroundings which are at 30 degree celsius.the specific heat values of liquid water and ice are 4.18 kj/kgk and 1.9 kj/kgk respectively and latent heat of fusion of water is 335 kj/kg.the minimum work in kj that is to be supplied to refrigerator to accomplish ice making is?
2. Relevant equations
efficiency= Q2/W = (T2/T1-T2)
H = mCpT
Q2=heat rejected by refrigerater
T2= temperature of ice (-18 degree celsius)
T1= temperature of surroundings

3. The attempt at a solution
don't know how to find mass of ice??the correct answer is 48.2kj
suggest me how to proceed

2. Mar 2, 2015

### Bystander

Step 1: read the problem statement.

3. Mar 2, 2015

### anisha

how shall i convert that,please explain in detail

4. Mar 3, 2015

### rude man

Start as follows:
For each of 3 separate steps,
what is the heat lost by the 1kg water/ice?

5. Jun 1, 2015

### VSURE

How you got 48.2 kJ? please give answer in step by step?