# How Much Water Can a 1KW Carnot Engine Freeze in 5 Minutes?

• zilex191
In summary, a 1KW electric motor drives a Carnot engine using an ideal gas to freeze water in 5 minutes. The mass of water frozen is 9.6 kg.
zilex191
Homework Statement
In a well-insulated refrigeration unit, a Carnot engine using an ideal gas is driven by a 1KW electric motor (80% efficient) to freeze water. Assuming that the temperature of the thermal sink is 20 °C, calculate the mass of water frozen in 5 minutes. Take the latent heat of water as 3.4 x 105 JKg-1
Relevant Equations
ncarnot = 1 -Qout/Qin = 1 - TL/TH
m = Q/L
W=Pt
Homework Statement: In a well-insulated refrigeration unit, a Carnot engine using an ideal gas is driven by a 1KW electric motor (80% efficient) to freeze water. Assuming that the temperature of the thermal sink is 20 °C, calculate the mass of water frozen in 5 minutes. Take the latent heat of water as 3.4 x 105 JKg-1
Homework Equations: ncarnot = 1 -Qout/Qin = 1 - TL/TH
m = Q/L
W=Pt

I've been stuck on this question for a long time ; here is what i have tried so far:

In a well-insulated refrigeration unit, a Carnot engine using an ideal gas is driven by a 1KW electric motor (80% efficient) to freeze water. Assuming that the temperature of the thermal sink is 20 °C, calculate the mass of water frozen in 5 minutes. Take the latent heat of water as 3.4 x 105 JKg-1

Formula for thermal efficiency:

ncarnot = 1 -Qout/Qin = 1 - TL/THQin i persume is given (20 °C) and ncarnot= 0.8 ; rearranging the formula you get Qout =100 °C

Wnet,out = Qout - Qin; Therefore Wnet,out = 80

at this point I'm lost, i know i need to find Q to fit into the formula m = Q/L as latent heat of water is given, but how do i combine the ncarnot with the latent heat formula+ time. Help will be much appreciated

The answer is supposed to be 9.6 kg

80% is the motor efficiency, not the Carnot cycle efficiency.

JD_PM
Ah ok thank you for your reply, so i use formula Output/input = 0.8, input into motor is 1000 W therefore output = 800 W

To work out energy transferred W=P x T ; W = 800 x (5minutes x 60) = 240000J

plug that into m= Q/L i get 0.71 kg.

Am i on the right track??

Heat is removed from the water to form ice at the cold reservoir temperature of 0 C and transferred to the ideal gas at 0 C;, and a greater amount of heat is rejected from the ideal gas to the hot reservoir (sink) at 20 C. The difference between these two amounts of heat is the work done.

## 1. What is a Carnot engine?

A Carnot engine is an idealized heat engine that operates on the Carnot cycle, which is a reversible thermodynamic process. It is used to understand the basic principles of thermodynamics and is often used as a theoretical benchmark for other real-life heat engines.

## 2. How does a Carnot engine work?

A Carnot engine works by taking in heat energy from a high-temperature reservoir, converting some of it into work, and then releasing the remaining heat energy into a low-temperature reservoir. It operates through a series of reversible processes, including isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

## 3. What is the efficiency of a Carnot engine?

The efficiency of a Carnot engine is given by the formula: efficiency = (Thigh - Tlow) / Thigh, where Thigh is the temperature of the high-temperature reservoir and Tlow is the temperature of the low-temperature reservoir. This means that the efficiency of a Carnot engine is always less than 1, and it increases as the temperature difference between the two reservoirs increases.

## 4. What are the limitations of a Carnot engine?

There are two main limitations of a Carnot engine. Firstly, it is an idealized model and does not account for practical factors such as friction and heat loss. Secondly, it requires the use of reversible processes, which are not always achievable in real-life systems.

## 5. How is the Carnot cycle related to the second law of thermodynamics?

The Carnot cycle is closely related to the second law of thermodynamics, which states that heat energy cannot spontaneously flow from a cold object to a hot object. The Carnot cycle demonstrates this principle by requiring the use of reversible processes, which do not violate the second law. This shows that the efficiency of a Carnot engine is the maximum possible efficiency for any heat engine operating between the same two temperatures.

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