Finding Final Temperature of Boiling Water and Potatoes

AI Thread Summary
To find the final temperature after adding potatoes to boiling water, the heat lost by the water equals the heat gained by the potatoes. The initial temperatures are 100°C for water and 30°C for potatoes, with specific heats of 4.2 kJ/kg°C and 3.4 kJ/kg°C, respectively. The initial approach led to an incorrect final temperature of 125°C, prompting a reevaluation of the temperature difference setup. Correctly applying the heat transfer equation with the right temperature assignments resulted in a more reasonable final temperature of approximately 85.125°C. The discussion emphasizes the importance of correctly interpreting the temperature changes in heat transfer calculations.
erok81
Messages
454
Reaction score
0

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...
 
Last edited:
Physics news on Phys.org
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top