Finding Final Temperature of Boiling Water and Potatoes

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Homework Help Overview

The discussion revolves around finding the final temperature of boiling water after adding potatoes, involving concepts of heat transfer and thermal equilibrium. The problem is situated within the context of thermodynamics, specifically focusing on heat exchange between two substances at different temperatures.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the heat transfer equations and the setup of the problem, questioning the arrangement of initial and final temperatures in their calculations. There is discussion about the implications of negative temperature changes and how they relate to heat gain or loss.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the assumptions behind their temperature assignments. Some have suggested alternative setups that yield more reasonable results, while others are clarifying the interpretation of temperature changes in the context of heat flow.

Contextual Notes

There is an emphasis on the assumption that no heat is lost to the environment, which is a critical aspect of the problem. Participants are also reflecting on the implications of the Zeroth Law of Thermodynamics in their reasoning.

erok81
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Homework Statement



I have some boiling water that I am adding potatoes to. I am asked to find the final temp after the potatoes have been added to the boiling water.

Tiw=100C (initial boiling water temp)
Tip=30C (initial room temp potatoes)
mp=0.5kg (mass of potatoes)
mw=1.5L or 1.5kg (mass of water)
cw=4.2kj/kg*C (specific heat of water)
cp=3.4kj/kg*C (specific heat of potatoes)

Homework Equations



C=\frac{Q}{\Delta T}
c=\frac{C}{m}

Since no heat is assumed to be lost, Qlost=Qgained

Subscript w is for water and p is potato

The Attempt at a Solution



Qlost=Qgained

cwmw(Tf-Tiw)=cpmp(Tf-Tip)

Both objects reach thermal equilibrium therefore Tf is the same.

After some algebra I end up with:

T_f = \frac{(c_w m_w T_iw) - (c_p m_p T_ip)}{(c_w m_w) - (c_p m_p)}

Then plugging in my given values I end up with 125C which is no way correct.

I am thinking my error might be how I am choosing my Ti/Tf values. I saw an example somewhere which had a couple of the values switched. Which didn't really make sense because as far as I know, it's always Tf-Ti?

Where am I making my error? If needed I can post my algebra steps. I did them a few times and got the same answer every time...
 
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However if I swap the temp values like an example I saw...

cwmw(Twi-Tf)=cpmp(Tf-Tpi)

After the algebra I end up with...

T_f = \frac{(c_p m_p T_{pi}) + (c_w m_w T_{wi})}{(c_w m_w) + (c_p m_p)}

I get 85.125C. Which is a lot more reasonable.


If that's the case, why do the temps get represented like that?
 
erok81 said:
Qlost=Qgained

cwmw(Tf-Tiw)=cpmp(Tf-Tip)

Tf-Tw<0 and TF-Tp>0. If you say that the water loses a negative amount of heat, it means it gains heat.

The water loses cwmw(Tw-Tf) heat. The potato gains cpmp(Tf-Tp) heat, and they are equal.

Or you can say that the heat transferred to a substance is cm(Tf-Ti) which can be either positive or negative and the sum of the transferred amounts of heat is zero, as heat is not lost.

ehild
 
So in this case ΔT isn't necessarily Tf-Ti like most Δ's. It's more for interpretation...if that makes sense.

I get what you are saying. For this problems the final and initial are "swapped" since the water has to lose heat at the end. If I do final - initial, the ΔT is negative and therefore gains heat, which isn't the case.

Therefore I need to look at what ΔT is doing and set it up from there? Or am I just reaching for a reason why they are swapped and this doesn't make sense?

Thanks for the reply!
 
Remember the Zeroth Law of Thermodynamics:

Heat flows from a body with a higher temperature to a body with a lower temperature.
 
erok81 said:
So in this case ΔT isn't necessarily Tf-Ti like most Δ's. It's more for interpretation...if that makes sense.

I get what you are saying. For this problems the final and initial are "swapped" since the water has to lose heat at the end. If I do final - initial, the ΔT is negative and therefore gains heat, which isn't the case.

Therefore I need to look at what ΔT is doing and set it up from there? Or am I just reaching for a reason why they are swapped and this doesn't make sense?

Thanks for the reply!

You need not think if you add up the terms cm(tf-ti) and make the sum equal to zero.


ehild
 

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