Finding Final Temperature of Boiling Water and Potatoes

[erok81]

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.

T_iw=100C (initial boiling water temp)
T_ip=30C (initial room temp potatoes)
m_p=0.5kg (mass of potatoes)
m_w=1.5L or 1.5kg (mass of water)
c_w=4.2kj/kg*C (specific heat of water)
c_p=3.4kj/kg*C (specific heat of potatoes)

Homework Equations

[itex]C=\frac{Q}{\Delta T}[/itex]
[itex]c=\frac{C}{m}[/itex]

Since no heat is assumed to be lost, Q_lost=Q_gained

Subscript w is for water and p is potato

The Attempt at a Solution

Q_lost=Q_gained

c_wm_w(T_f-T_iw)=c_pm_p(T_f-T_ip)

Both objects reach thermal equilibrium therefore T_f is the same.

After some algebra I end up with:

[itex]T_f = \frac{(c_w m_w T_iw) - (c_p m_p T_ip)}{(c_w m_w) - (c_p m_p)}[/itex]

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 T_i/T_f 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 T_f-T_i?

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...

 

[erok81]

However if I swap the temp values like an example I saw...

c_wm_w(T_wi-T_f)=c_pm_p(T_f-T_pi)

After the algebra I end up with...

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

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


If that's the case, why do the temps get represented like that?

 

[ehild]

Q_lost=Q_gained

c_wm_w(T_f-T_iw)=c_pm_p(T_f-T_ip)

T_f-T_w<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 c_wm_w(Tw-Tf) heat. The potato gains c_pm_p(Tf-Tp) heat, and they are equal.

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

ehild

 

[erok81]

So in this case ΔT isn't necessarily T_f-T_i 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!

 

[SteamKing]

Remember the Zeroth Law of Thermodynamics:

Heat flows from a body with a higher temperature to a body with a lower temperature.

 

[ehild]

So in this case ΔT isn't necessarily T_f-T_i 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