Thermo, Rate of Heat transfer

  • Thread starter digipony
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  • #1
digipony
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Homework Statement


A cube of Al (.1m x .1m x .1m with mass=.27kg) is placed on a large thermal mass maintained at constant 40 C (313 K).The cube is separated from the thermal mass by a .01m thick glass plate (.1m x .1m). If the Al block is initially at 10 C, how long does it take to reach a temperature of 20 C?
(Assume that no heat is transferred to the enviornment and that the thermal mass of the glass plate can be ignored.) Hint: Determine the rate at which the heat is transferred for a given temperature of the Al cube and then determine the rate at which the temperature of the Al cube is changing. The total time is obtained by integrating your equation given the intial and final temps.


Homework Equations


Q=mcΔT
H=dQ/dt=(kAΔT)/L


The Attempt at a Solution


I worked with my professor, and thought I understood it, but now I am stuck. -_- Here is what I have done so far:

dQ/dt = mc dT/dt=kA[T_hot - T_al(t)]/L (T is temp, t is time, L is length)

mcL dT = kA[T_hot - T_al(t) ] dt

mcL(20-10) = kA(T_hot*t_20)-kA*integral[T_al(t)] (the integral is indefinite and from 0 to T_20,T_20 being what I am solving for)

10*mcl/kA=T_hot*t_20-integral[T_al(t)]

10*.27*910*.1/(.1*.1*205)= (313 Kelvin)*t_20-integral[T_al(t)]

119.85=(313 Kelvin)*t_20-integral[T_al(t)]

and here I am stuck, as I don't know what to do with the last integral on the right side. I think I need to find an equation for T_al(t) that relates t to T_al(t), however I cannot figure this out. I know that when t=0,T_al=10 C, but when t=?, T_al = 20 C.

Thanks! :smile:

**If anyone noticed that the work kept getting changed, It was because I am still trying to work on it while awaiting a reply and wanted to update the post as to what work I have done. Sorry for any confusion/inconvenience.
 
Last edited:

Answers and Replies

  • #2
22,332
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In your first equation, T_al(t) and T are the same parameter. T is the temperature of the block at time t. The heating rate is proportional to the difference between T_hot and T. Now you have the time derivative of T on the LHS, and a linear function of T on the RHS.
 
  • #3
digipony
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I don't quite follow. In the first equation, are you referring to dT/dt as T? And what about the integral?
 
  • #4
22,332
5,205
I don't quite follow. In the first equation, are you referring to dT/dt as T? And what about the integral?

The starting equation should read:

mc dT/dt=kA[T_hot - T]/L

I assume you have had a course in differential equations. You need to solve this differential equation subject to the initial condition

T = 10 @ t = 0

dT/[T_hot - T] = KAdt/(Lmc)
 
  • #5
digipony
35
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The starting equation should read:

mc dT/dt=kA[T_hot - T]/L

I assume you have had a course in differential equations. You need to solve this differential equation subject to the initial condition

T = 10 @ t = 0

dT/[T_hot - T] = KAdt/(Lmc)

I am in Differential equations right now, and I did not realize that this is the method to use, but I see where I need to go from here. Thank you so much for your help!
 

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