In a series of experiments, block Bis to be placed in a thermally insulated container with block A, which has the same mass as block B. In each experiment, block Bis initially at a certain temperature${\mathit{T}}_{\mathbf{B}}$, but temperature ${\mathit{T}}_{\mathbf{A}}$of block Ais changed from experiment to experiment. Let ${\mathit{T}}_{\mathbf{f}}$ represent the final temperature of the two blocks when they reach thermal equilibrium in any of the experiments.

Figure 18-53 gives temperature${\mathit{T}}_{\mathbf{f}}$ versus the initial temperature ${\mathit{T}}_{\mathbf{A}}$ for a range of possible values of ${\mathit{T}}_{\mathbf{A}}$, from ${\mathit{T}}_{\mathbf{A}\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{\text{K}}$ to${\mathit{T}}_{\mathbf{A}\mathbf{2}}\mathbf{=}\mathbf{500}\mathbf{}\mathbf{\text{K}}$. The vertical axis scale is set by ${\mathit{T}}_{\mathbf{f}}\mathbf{=}\mathbf{400}\mathbf{}\mathbf{\text{K}}$. What are (a) temperature${\mathit{T}}_{\mathbf{B}}$ and (b) the ratio $\raisebox{1ex}{${\mathbf{C}}_{\mathbf{B}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{C}}_{\mathbf{A}}$}\right.$of the specific heats of the blocks?

Figure 18.53

1. Block B is placed in the thermally insulated container with block A.
2. Mass of block B ($m_B$) = Mass of block A ($m_A$)
3. Initial temperature of block B is $T_B$.
4. Initial temperature of block A is $T_A$.
5. Final temperature of both blocks at equilibrium is $T_f$ .
6. vs. graph is given.

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. Since the system is insulated we can equate initial and final heat energy of the system in terms of specific heat and temperature of block A and B. Rearranging the equation we can find an equation for the given graph. From this we can get the expressions for the slope and an intercept. Inserting the values of slope and an intercept , we can find the temperature of block B and the ratio of the specific heat of the blocks.

Formula:

The heat absorbed or transferred by the body,

$Q=mc\Delta T$ ......(i)

Since the system is insulated, the heat transferred by block A is equal to the heat transferred by block B.

$Q_A=Q_B$

From equation (i), we can write the heat values as,

$\begin{array}{rcl}{m}_A{c}_A{T}_A+m_B{c}_B{T}_B& =& m_A{c}_A{T}_f+m_B{c}_B{T}_f\\ c_A{T}_A+c_B{T}_B& =& c_A{T}_f+c_B{T}_f\text{(}\because m_A=m_B\text{)}\\ T_f& =& \frac{c_A{T}_A+c_B{T}_B}{c_A+c_B}\\ & =& \frac{c_A}{c_A+c_B}{T}_A+\frac{c_B}{c_A+c_B}{T}_B\end{array}$

It is a straight line equation having

$slope=\frac{c_A}{c_A+c_B}$and $intercept=\frac{c_B}{c_A+c_B}{T}_B$

$$\begin{gathered} T_B=intercept\left(\frac{c_A+c_B}{c_B}\right) \\ {T}_B=\frac{intercept}{1-slope} \end{gathered}$$

From the graph we can infer that, Intercept is $150\text{K}$and slope is $\frac{2}{5}$

From the equation (ii) and above values, we can say that the temperature of the block B is

$\begin{array}{rcl}{T}_B& =& \frac{150 \text{K}}{\left(1-\frac{2}{5}\right)}\\ & =& \frac{150}{\left(\frac{3}{5}\right)}\\ & =& 250\text{K}\\ & =& 2.5\times {10}^2\text{K}\end{array}$

Therefore, the temperature of block B (T_B) is .$2.5\times {10}^2\text{K}$

From equation (ii) of part (a), we can write that the ratio of the specific heats is

$\begin{array}{rcl}{T}_B& =& \text{intercept}\left(\frac{c_A+c_B}{c_B}\right)\\ T_B& =& \text{intercept}\left(\frac{c_A}{c_B}+1\right)\\ \frac{c_A}{c_B}& =& \frac{T_B}{\text{intercept}}-1\\ & =& \frac{250}{150}-1\\ & =& \frac{2}{3}\\ \frac{c_B}{c_A}& =& \frac{3}{2}\\ & =& 1.5\end{array}$

Therefore, the ratio of the specific heats of the blocks is $1.5$