Estimate the mass of the water (Carnot cycle)

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SUMMARY

The discussion focuses on estimating the mass of water that can be heated from 20°C to 100°C using a Carnot heat pump powered by a phone battery with a voltage of 3.6V and a capacity of 2800 mAh. The key equation derived is m=Q/c(79*ln80), where Q represents the heat energy transferred. Participants emphasized the need to account for the time the battery operates to determine the total energy delivered, calculated as voltage multiplied by capacity and time. The conversation also touched on the change in entropy for both the water and the surroundings during the heating process.

PREREQUISITES
  • Understanding of the Carnot cycle and its application in heat pumps
  • Knowledge of thermodynamic principles, specifically entropy and heat transfer
  • Familiarity with electrical concepts, including voltage and capacity in mAh
  • Basic calculus for integrating temperature-dependent coefficients
NEXT STEPS
  • Research the principles of the Carnot cycle and its efficiency calculations
  • Learn about the relationship between voltage, current, and energy in electrical systems
  • Study the concept of entropy changes in thermodynamic processes
  • Explore numerical integration techniques for temperature-dependent functions
USEFUL FOR

Students in thermodynamics, electrical engineering enthusiasts, and anyone involved in energy transfer calculations using heat pumps.

fizzyfiz
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Homework Statement
Given voltage and current, estimate how much mass of water can be heated from 20 to 100 degrees celsius by carnot heat pump. No heat losses. I came up with the idea of equation below, temperature is changing, so is efficiency. I sum multiples of heat required to increase tempreture by dT and efficiency at that T but I do not have any idea how to involve energy supplied into calculations.

I used "S" here to indicate integral.
Relevant Equations
Q=(293, 373)S T/T-293 *cm dT
m=Q/c(79*ln80)
 
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fizzyfiz said:
Homework Statement: Given voltage and current, estimate how much mass of water can be heated from 20 to 100 degrees celsius by carnot heat pump. No heat losses. I came up with the idea of equation below, temperature is changing, so is efficiency. I sum multiples of heat required to increase tempreture by dT and efficiency at that T but I do not have any idea how to involve energy supplied into calculations.

I used "S" here to indicate integral.
Homework Equations: Q=(293, 373)S T/T-293 *cm dT

m=Q/c(79*ln80)
Please provide the exact word-for-word statement of the problem.
 
Estimate how much mass can be heated from 20( same as surroudnigs) to 100 degress celsius using phone battery of volatge 3.6 and capacity of 2800 mAh using heat pump. Heat pump must be treated as carnot cycle backwards.

So I know that the coefficient is changing as temperature is changing. The energy delivered to the water is sum of all coeficients multiplied by the energy to heat the mass of water by 1K. I am struggling to write the sum properly.
 
fizzyfiz said:
Estimate how much mass can be heated from 20( same as surroudnigs) to 100 degress celsius using phone battery of volatge 3.6 and capacity of 2800 mAh using heat pump. Heat pump must be treated as carnot cycle backwards.

So I know that the coefficient is changing as temperature is changing. The energy delivered to the water is sum of all coeficients multiplied by the energy to heat the mass of water by 1K. I am struggling to write the sum properly.
Do they give you any indication of how long the battery is delivering the power to run the heat pump?
 
Chestermiller said:
Do they give you any indication of how long the battery is delivering the power to run the heat pump?
No they do not. I think that I should assume that the whole energy is transfered.
 
fizzyfiz said:
No they do not. I think that I should assume that the whole energy is transfered.
To get the whole energy, you have to specify the amount of time the battery is delivering power. You can express the answer to this question as a function of t, the time of operation.
 
The energy delivered by the battery is equal to volatage*2800 mA*3600s *10^-3. Then it is trasfered to water via backward Carnot cycle.
 
fizzyfiz said:
The energy delivered by the battery is equal to volatage*2800 mA*3600s *10^-3. Then it is trasfered to water via backward Carnot cycle.
Where in your problem statement does it say anything about an hour?
 
They does not but I am given capacity of battery in mAh. So I multiply it by 3600 s to get charge in columbs.
 
  • #10
fizzyfiz said:
They does not but I am given capacity of battery in mAh. So I multiply it by 3600 s to get charge in columbs.
Oh. OK. I missed that.

OK. Let M represent the mass of the water and C represent the heat capacity of the water. In terms of M and C, what is the change in entropy of the water in going from 20 C to 100 C? In terms of M and C, how much heat Q is added to the water in going from 20 C to 100 C?

If the heat pump is operated reversibly, what is the change in entropy of the surroundings (cold reservoir)?
 
  • #11
I managed to do it by myself yesterday, thank you for your help :)
 

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