What is the average power of the warmer bath in this thermal dynamics problem?

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Homework Help Overview

The discussion revolves around a thermal dynamics problem involving heat transfer between a warm water bath and an ice-water bath through a metal rod. The original poster presents equations related to power and temperature change over time, seeking clarification on their approach and the implications of Fourier's law in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between power and temperature difference, questioning the original poster's interpretation of the equations. There are attempts to clarify the distinction between time and temperature variables in the equations presented.

Discussion Status

There is ongoing dialogue about the correct application of Fourier's law and the formulation of the differential equations involved. Some participants have offered insights into potential misunderstandings and suggested alternative formulations, while others express confusion about the problem setup and the calculations involved.

Contextual Notes

Participants note that certain details from the problem statement may have been overlooked, which could affect the understanding of the equations. There is also mention of the need to consider average temperatures in the context of calculating average power, highlighting the complexity of the problem.

MMONISM
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Homework Statement


A warm water bath containing 10.0L of water is connected to an ice-water bath with a piece of metal of length L = 2.11 m and cross sectional area A = 1975 cm2. The metal has a thermal conductivity of km = 60.5 Wm-1K-1, a specific heat of cm = 239.7 Jkg-1K-1 and a density of ρm = 3782.5 kgm-3.The warm water bath is initially at a temperature of Th = 63.6 °C.
a413cff7gw1esalixwpooj20vg0aidgz.jpg

if the power source heating the warm water bath is switched off. In this case the temperature of the warm water bath will gradually decrease as heat is transferred to the cool water bath. We can describe the heat lost by the warm water bath and the metal rod (the average temperature is just the averages of the temperatures on either side) in time tf as:
##P(t) = -m_wc_w\frac{dT_f}{dt} -\frac{m_mc_m}{2}\frac{dT_f}{dt}##
In this equation the final temperature and power are functions of tf, the other variables are not dependent on tf. We can then differentiate with respect to time to get the expression (replacing tf with t here):
##P(t) = -m_wc_w\frac{dT_f}{dt} -\frac{m_mc_m}{2}\frac{dT_f}{dt}##
which be rearranged to give:
##\frac{dT_f}{dt} = -\frac{P}{m_wc_w+m_mc_m/2}##
what is the temperature of the warm water bath after one hour has passed.

Homework Equations


##\frac{dT_f}{dt} = -\frac{P}{m_wc_w+m_mc_m/2}##
##P = KA \frac{Th - Tc}{L}##

The Attempt at a Solution


a413cff7gw1esamhzo1o9j23gg22ohdu.jpg

Could someone tell me if my approach is correct? Thanks in advance.
 
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Hi, I see you get little response.

Here's something: P doesn't only depend on time, it also depends on the temperature difference between the baths (Fourier's law). So you get a different differential equation, I'm afraid.
 
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BvU said:
Hi, I see you get little response.

Here's something: P doesn't only depend on time, it also depends on the temperature difference between the baths (Fourier's law). So you get a different differential equation, I'm afraid.

Hi, I am really appreciate for your reply. but I am not sure what do you mean.
because I think I used Fourier's law also I am sorry for i didn't make myself clear as ##\frac{dT_f}{dt} = -\frac{P}{m_wc_w+m_mc_m/2}## is given by the question. therefore I think I have to combine these two formula as what I have done in the picture. finally, do you think my approach is reasonable?
a413cff7gw1esb5wy5pgjj22p81mc7wk.jpg
 
Could you distinguish tf and Tf ? one is a time, the other a temperature.

If ## {dT(t) \over dt} = \alpha T(t)## that is a differential equation, not an algebraic one.

MMONISM said:
We can describe the heat lost by the warm water bath and the metal rod (the average temperature is just the averages of the temperatures on either side) in time tf as:
$$P(t) = -m_wc_w\frac{dT_f}{dt} -\frac{m_mc_m}{2}\frac{dT_f}{dt}$$
In this equation the final temperature and power are functions of tf
There is no tf in this whole expression ?! And if the dimensions match depend on what you think P(t) is . It certainly isn't heat.
 
BvU said:
Could you distinguish tf and Tf ? one is a time, the other a temperature.

If ## {dT(t) \over dt} = \alpha T(t)## that is a differential equation, not an algebraic one.

There is no tf in this whole expression ?! And if the dimensions match depend on what you think P(t) is . It certainly isn't heat.
I kind of understand what do you mean now, but I am so confused about the question too, that's why I want some tips for the question, but still really appreciate for your reply. if you think the picture below can help you to figure out your confusion and you do have some ideal how to work out the last question, please give me some tips about the question(as the picture shows I tried my approach with answer 7.5 hrs, but it is not the correct answer).
捕获.PNG
 
Aha, there were a few chunks from the problem statement that you left out (without malicious intent, I am sure) and indeed resolve my confusion :smile: !
And you should be able to understand the difference between the actual problem and the rendering as I quoted in post #4 ! The text is about the equation for Q you accidentally replaced by the equation for P.

It also dawns on me that, to answer (b), you effectively use the value for P at t=1 hour (the formula as found in part (a)) -- which I think is very crooked.

It only works because P is small enough. If you would have used the value for P at t=0 that would have given 57.98 degrees, which also would have been found OK by the program, I suppose.

The correct way (I think) would be to write $$
{dT(t)\over dt} = {k_m A\over L} \ {T(t) - 0 ^\circ{\rm C}\over m_wc_w + m_m c_m/2}$$giving $$
{dT(t)\over T} = {k_m A\over L} \ {1\over m_wc_w + m_m c_m/2}\; dt$$
which you can solve, I suppose.
 
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BvU said:
Aha, there were a few chunks from the problem statement that you left out (without malicious intent, I am sure) and indeed resolve my confusion :smile: !
And you should be able to understand the difference between the actual problem and the rendering as I quoted in post #4 ! The text is about the equation for Q you accidentally replaced by the equation for P.

It also dawns on me that, to answer (b), you effectively use the value for P at t=1 hour (the formula as found in part (a)) -- which I think is very crooked.

It only works because P is small enough. If you would have used the value for P at t=0 that would have given 57.98 degrees, which also would have been found OK by the program, I suppose.

The correct way (I think) would be to write $$
{dT(t)\over dt} = {k_m A\over L} \ {T(t) - 0 ^\circ{\rm C}\over m_wc_w + m_m c_m/2}$$giving $$
{dT(t)\over T} = {k_m A\over L} \ {1\over m_wc_w + m_m c_m/2}\; dt$$
which you can solve, I suppose.
OMG, that was a huge mistake:nb). really really sorry for that mistake I made.

BvU said:
The correct way (I think) would be to write
dT(t)dt=kmAL T(t)−0∘Cmwcw+mmcm/2​
{dT(t)\over dt} = {k_m A\over L} \ {T(t) - 0 ^\circ{\rm C}\over m_wc_w + m_m c_m/2}giving
dT(t)T=kmAL 1mwcw+mmcm/2dt​
{dT(t)\over T} = {k_m A\over L} \ {1\over m_wc_w + m_m c_m/2}\; dt
which you can solve, I suppose.
But for the second question what would T be? average temperature Tf and Ti? because function T (t) is not given and same problem confused me at the third question.

P50522-005306(1).jpg

if Tf is half of Ti, during the process I can't use Fourier's law as it only calculate the instantaneous power, but if i want to know dt(tf - 0) I need to find the average power by the warmer bath, which as my approach shows I used the average temperature of Tf and Ti and final answer is 7.5 which is not correct as you said before. but how am I be able to find out the average power by the warmer bath throughout the whole process? and once again really thanks your patience and your understanding.:smile:
 

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