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Estimate the mass of the water (Carnot cycle)

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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.
Homework Equations
Q=(293, 373)S T/T-293 *cm dT
m=Q/c(79*ln80)
 
19,345
3,836
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.
 
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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.
 
19,345
3,836
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?
 
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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.
 
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3,836
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.
 
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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.
 
19,345
3,836
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?
 
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They does not but I am given capacity of battery in mAh. So I multiply it by 3600 s to get charge in columbs.
 
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3,836
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)?
 
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I managed to do it by myself yesterday, thank you for your help :)
 

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