How is the slope of the temperature graph determined?

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ShizukaSm
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In a series of experiments, block B is to be placed in a thermally insulated container with
block A, which has the same mass as blockB. In each experiment, block B is initially at a certain temperature TB, but temperature TA of block A is changed from experiment to experiment. Let Tf
represent the final temperature of the two blocks when they reach thermal equilibrium in any of the experiments. The graph(attatched) gives temperature Tf versus the initial temperature TA for a range of possible values of TA, from TA = 0 K to TA = 500 K. The vertical axis scale is set by Tfs= 400 K. What are:
(a)temperature TB.
(b) the ratio cB/cA of the specific heats of the blocks?
sfa.JPG


Ok so, I was able to solve this problem, however, my book answer used a method that I did not understand:
Slope.JPG


How can he infer that the slope is equal to [itex]\frac{c_A}{c_A+c_B}[/itex]? Where did that come from?
 
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ShizukaSm said:
In a series of experiments, block B is to be placed in a thermally insulated container with
block A, which has the same mass as blockB. In each experiment, block B is initially at a certain temperature TB, but temperature TA of block A is changed from experiment to experiment. Let Tf
represent the final temperature of the two blocks when they reach thermal equilibrium in any of the experiments. The graph(attatched) gives temperature Tf versus the initial temperature TA for a range of possible values of TA, from TA = 0 K to TA = 500 K. The vertical axis scale is set by Tfs= 400 K. What are:
(a)temperature TB.
(b) the ratio cB/cA of the specific heats of the blocks?
View attachment 60477

Ok so, I was able to solve this problem, however, my book answer used a method that I did not understand:
View attachment 60478

How can he infer that the slope is equal to [itex]\frac{c_A}{c_A+c_B}[/itex]? Where did that come from?
Rewriting the expression as ##T_f=\left(\frac{c_A}{c_A+c_B}\right) T_A+\left(\frac{c_B}{c_A+c_B}\right) T_B## might help. :wink:
 
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Mandelbroth said:
Rewriting the expression as ##T_f=\left(\frac{c_A}{c_A+c_B}\right) T_A+\left(\frac{c_B}{c_A+c_B}\right) T_B## might help. :wink:

Oh, yes it does! Thanks a lot:smile:
 
ShizukaSm said:
Oh, yes it does! Thanks a lot:smile:
You're most certainly welcome. Good luck with the physics. :wink: