Temperature graph slope

[ShizukaSm]

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?

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

How can he infer that the slope is equal to [itex]\frac{c_A}{c_A+c_B}[/itex]? Where did that come from?

 

[Mandelbroth]

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.

 

[ShizukaSm]

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.

Oh, yes it does! Thanks a lot

 

[Mandelbroth]

Oh, yes it does! Thanks a lot

You're most certainly welcome. Good luck with the physics.

 

[Nickie]

The slope of a graph represents the rate of change between two variables. In this case, the slope of the graph represents the change in temperature (Tf) with respect to the change in initial temperature (TA). We can use this information to find the ratio of the specific heats of the two blocks.

The specific heat of a substance is the amount of heat required to raise the temperature of one unit mass of the substance by one degree. In this case, we have two blocks (A and B) with the same mass, so we can assume that they have the same specific heat (cA = cB).

Now, in each experiment, block B is initially at a certain temperature TB, but the temperature of block A (TA) is changed. This means that the change in temperature for block A (ΔTA) is equal to the change in temperature for block B (ΔTB).

Using the slope of the graph, we can write the following equation:

Slope = ΔTf/ΔTA = ΔTf/(ΔTA + ΔTB)

Since we know that ΔTA = ΔTB, we can rewrite the equation as:

Slope = ΔTf/2ΔTA

Now, we also know that the final temperature (Tf) is equal to the average of the initial temperatures of block A and B (TA and TB), so we can write:

Tf = (TA + TB)/2

Substituting this into the previous equation, we get:

Slope = (Tf - TB)/ΔTA

Finally, we know that the vertical axis scale is set by Tf = 400 K, so we can write:

Slope = (400 K - TB)/ΔTA

Rearranging this equation, we get:

TB = 400 K - Slope x ΔTA

Now, we can see that the slope is equal to the ratio of the specific heats of the blocks, since the specific heats are the only variables in the equation. Therefore, we can write:

TB = 400 K - (cB/cA) x ΔTA

This is how we can infer that the slope is equal to cB/cA. I hope this helps!