Estimate the mass of the water (Carnot cycle)

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Homework Help Overview

The discussion revolves around estimating the mass of water that can be heated from 20 to 100 degrees Celsius using a Carnot heat pump powered by a phone battery. Participants explore the relationship between energy supplied, temperature changes, and efficiency in the context of thermodynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to account for changing efficiency as temperature varies and express uncertainty about how to incorporate energy supplied into their calculations. There are attempts to formulate the energy transfer and its relation to the mass of water being heated.

Discussion Status

The conversation includes various interpretations of the problem, particularly regarding the assumptions about battery operation time and energy transfer. Some participants suggest expressing the energy delivered as a function of time, while others clarify the relationship between battery capacity and energy calculations.

Contextual Notes

Participants note the lack of explicit information about the duration of battery operation, leading to assumptions about energy transfer. The problem's constraints include the requirement to treat the heat pump as a Carnot cycle operating in reverse.

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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