How Much Water Vapor Is Needed to Heat a Pot from 30C to 60C?

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Homework Help Overview

The problem involves determining the mass of water vapor needed to heat a pot containing water from 30°C to 60°C. The setup includes 2 kg of water and a 0.5 kg aluminum pot, with specific heat capacities provided for both water and aluminum.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the heat transfer equations and the implications of using water vapor at 100°C to heat the pot and water. There are questions about the correct formulation of the heat transfer equation and whether to account for the condensation of the vapor.

Discussion Status

Participants are exploring different interpretations of the problem, including the role of heat from the vapor and the conditions under which thermal equilibrium is achieved. Some have suggested that the problem may be poorly worded, leading to multiple reasonable interpretations regarding the final temperature and heat transfer processes involved.

Contextual Notes

There are discussions about the assumptions regarding thermal equilibrium and the potential constraints imposed by the second law of thermodynamics, particularly in relation to the temperature of the condensate.

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Homework Statement


There are 2 kg of water in an aluminium pot at the temperature of 30C. What is the mass of water vapor (100C) needed to heat the pot with water to 60C? Mass of the pot = 0.5kg

m(water)=2kg
m(pot)=0.5kg
T1=30C
T2=60C
c(water)=4200 J/kg*C
c(aluminium)=890 J/kg*C
T(water vapor)=100*C
c(water vapor)= 2020 J/kg*C
m(water vapor)=?

2. The attempt at a solution

Q=Q1+Q2
Q=c(water)*m(water)*(T2-T1)+c(aluminium)*m(aluminium)*(T2-T1)

I supposed that
Q=c(vapor)*m(vapor)*(T2-T(vapor))
But then Q would be less than 0 since the temperature difference is less than 0. What is the right way to write this last equation? And the final equation to find the mass of water vapor?
Thank you!
 
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You write Q=c(vapor)*m(vapor)*(T2-T(vapor)), but the formulation says What is the mass of water vapor (100 C), not taking account of the condensation of vapor water. This is, we use m Kg of vapor water to heat up the Al pot, which transfer heat to the water inside it. I'm thinking more of Q=c(vapor)*m(vapor)*(100)

On the other hand, you are supposing that both Al pot and the water inside it will reach the same temperature (T2). That would be the case according to zeroth law of thermodynamics, in thermal equilibrium (for large t, time). One could put it like: 1) Final T2 in the pot for the water to reach 60c? and 2) mass of vapor water to heat the pot to that T2 calculated in 1. This is my guess...
 
At best, this is a poorly (imprecisely) worded problem. I think what they are asking is how much water vapor has to condense from vapor to liquid water at 100 C to supply the heat required to heat up the pot and its contents to 60C. So, how much heat does it take to heat up the contents?

Chet
 
Chestermiller said:
At best, this is a poorly (imprecisely) worded problem. I think what they are asking is how much water vapor has to condense from vapor to liquid water at 100 C to supply the heat required to heat up the pot and its contents to 60C. So, how much heat does it take to heat up the contents?

Chet
Maybe, but there are at least two other reasonable interpretations. There is still useful heat available in condensate at 100C, so could take everyting as finishing at 60C. Better still, the initial condensate is useful down to almost 30C, so the minimum vapor required is less again.
 
haruspex said:
Maybe, but there are at least two other reasonable interpretations. There is still useful heat available in condensate at 100C, so could take everyting as finishing at 60C. Better still, the initial condensate is useful down to almost 30C, so the minimum vapor required is less again.
Yes. I too thought about the possibility of taking the condensate down to 60, and this is certainly also a valid interpretation. As far as taking the condensate all the way down to 30, it doesn't seem like this can be done because of 2nd law constraints. Can you think of a way?

Chet
 

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