# Specific Heat Problem

1. Jan 21, 2013

### erok81

1. The problem statement, all variables and given/known data

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)

2. Relevant 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

3. 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...

Last edited: Jan 21, 2013
2. Jan 21, 2013

### erok81

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?

3. Jan 21, 2013

### ehild

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

4. Jan 21, 2013

### erok81

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?

5. Jan 22, 2013

### SteamKing

Staff Emeritus
Remember the Zeroth Law of Thermodynamics:

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

6. Jan 22, 2013

### ehild

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

ehild