Thermodynamics-reversible carnot engine

AI Thread Summary
The discussion revolves around calculating the minimum work required for a refrigerator to convert 1 kg of liquid water at 30 degrees Celsius to ice at -18 degrees Celsius. Key thermodynamic principles are applied, including the specific heat capacities of water and ice, as well as the latent heat of fusion. The efficiency of the refrigerator is defined using the temperatures of the ice and surroundings. Participants seek clarification on calculating the heat lost during the phase changes and the overall energy required, ultimately arriving at a solution of 48.2 kJ. Detailed step-by-step guidance is requested to understand the calculations involved in this thermodynamic process.
anisha
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Homework Statement


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?

Homework 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

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
 
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anisha said:
don't know how to find mass of ice?
anisha said:
convert 1 kg of liquid water at 30 degree celsius to ice
Step 1: read the problem statement.
 
how shall i convert that,please explain in detail
 
Start as follows:
For each of 3 separate steps,
what is the heat lost by the 1kg water/ice?
 
How you got 48.2 kJ? please give answer in step by step?
 
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