Question

A 364 gblock is put in contact with a thermal reservoir. The block is initially at a lower temperature than the reservoir. Assume that the consequent transfer of energy as heat from the reservoir to the block is reversible. Figure gives the change in entropy $\mathbf{∆}\mathit{S}$ of the block until thermal equilibrium is reached. The scale of the horizontal axis is set by${\mathit{T}}_{\mathbf{a}}\mathbf{=}\mathbf{280}\mathit{K}\mathbf{}\mathbf{and}\mathbf{}{\mathit{T}}_{\mathbf{b}}\mathbf{=}\mathbf{380}\mathit{K}$. What is the specific heat of the block?

Short answer

Specific heat of the block is 450 J/kg K .

Step-by-step solution

The given data

a) Mass of the block, m = 364 g or 0.364 kg

b) The graph of entropy change vs. temperature is given.

c) Temperature of horizontal axis, $T_a=280K\mathrm{and}{T}_b=380K$

Understanding the concept of entropy change

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We can write the formula for specific heat by rearranging the formula for entropy change. Then inserting the values obtained from the given graph, we can find thespecific heat of the block.

Formula:

The entropy change of the gas, $∆S=mc\ln \left(\frac{T_f}{T_i}\right)$ …(i)

Step 3: Calculation for the specific heat of the block

Using equation (i) and the given values, the specific heat of the block is given as:

(From the graph, we can infer that

$T_f=380K,T_i=280K\mathrm{and}∆\mathrm{S}=\mathrm{J}/\mathrm{K}.$)

$$\begin{gathered} c=\frac{∆S}{mln\left({\displaystyle \frac{T_f}{T_i}}\right)} \\ =\frac{50\mathrm{J}/\mathrm{K}}{\left(0.364\mathrm{kg}\right)In\left({\displaystyle \frac{380K}{280K}}\right)} \\ =449.8\mathrm{J}/\mathrm{kgK} \\ ~450\mathrm{J}/\mathrm{kgK} \end{gathered}$$

Hence, the value of the specific heat of the block is $450\mathrm{J}/\mathrm{kgK}$