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

In summary, the problem involves a warm water bath connected to an ice-water bath through a metal rod, with given values for thermal conductivity, specific heat, and density. The warm water bath starts at 63.6 °C and after the power source is switched off, the temperature gradually decreases. The heat lost by the warm water bath and the metal rod can be described using a differential equation, and the temperature of the warm water bath after one hour can be solved for using this equation. However, the given formula for power was incorrectly used in the original attempt, resulting in an incorrect answer. The correct approach would involve a different formula for power and solving for temperature using a differential equation.
  • #1
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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  • #2
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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  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
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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  • #7
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:
 

1. What is thermal dynamics?

Thermal dynamics is the study of heat and its relationship with other forms of energy, such as work and mass. It involves understanding how heat is transferred and transformed in various systems and processes.

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