Thermodynamics-reversible carnot engine

In summary, the problem involves converting 1 kg of liquid water at 30 degree celsius to ice at -18 degree celsius using a refrigerator that rejects heat to the surroundings 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 the latent heat of fusion of water is 335 kj/kg. The minimum work in kj that needs to be supplied to the refrigerator to accomplish ice making is 48.2kj. To find the mass of ice, we need to consider three separate steps and calculate the heat lost in each step. The final answer can be obtained by adding up the heat lost in each step.
  • #1
anisha
9
0

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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  • #2
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.
 
  • #3
how shall i convert that,please explain in detail
 
  • #4
Start as follows:
For each of 3 separate steps,
what is the heat lost by the 1kg water/ice?
 
  • #5
How you got 48.2 kJ? please give answer in step by step?
 

FAQ: Thermodynamics-reversible carnot engine

What is a reversible Carnot engine?

A reversible Carnot engine is a theoretical engine that operates between two heat reservoirs and produces maximum work output. It follows the Carnot cycle, which consists of two isothermal processes and two adiabatic processes.

What is the efficiency of a reversible Carnot engine?

The efficiency of a reversible Carnot engine is given by the equation: efficiency = (Thot - Tcold) / Thot, where Thot is the temperature of the hot reservoir and Tcold is the temperature of the cold reservoir. This means that the efficiency of a Carnot engine increases with higher temperature difference between the two reservoirs.

How does a reversible Carnot engine compare to a real engine?

A reversible Carnot engine is a theoretical ideal engine that operates with maximum efficiency. In reality, engines are not perfectly efficient and have losses due to friction, heat transfer, and other factors. However, the Carnot engine serves as a useful benchmark for comparing the performance of real engines.

Can the efficiency of a reversible Carnot engine be improved?

No, the efficiency of a reversible Carnot engine cannot be improved beyond the maximum theoretical efficiency. This is because it operates on a reversible cycle, where there is no irreversibility or energy losses. Any real engine will have some level of irreversibility, leading to lower efficiency.

What are the applications of a reversible Carnot engine?

A reversible Carnot engine is used as a theoretical model to understand the fundamental principles of thermodynamics. It also serves as a benchmark for the efficiency of real engines. Additionally, the Carnot engine concept is used in refrigerators and heat pumps, which are essentially reversed Carnot engines that transfer heat from a cold reservoir to a hot reservoir.

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