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Q10P

Expert-verifiedFound in: Page 604

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A 364 g**** block 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 ${\mathbf{\u2206}}{\mathit{S}}$** **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?**

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

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, ${T}_{a}=280K\mathrm{and}{T}_{b}=380K$

**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 the ****specific heat of the block.**

Formula:

The entropy change of the gas, $\u2206S=mc\mathrm{ln}\left(\frac{{T}_{f}}{{T}_{i}}\right)$ …(i)

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

(From the graph, we can infer that

${T}_{f}=380K,{T}_{i}=280K\mathrm{and}\u2206\mathrm{S}=\mathrm{J}/\mathrm{K}.$ )

$c=\frac{\u2206S}{mln\left(\frac{{T}_{f}}{{T}_{i}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{50\mathrm{J}/\mathrm{K}}{\left(0.364\mathrm{kg}\right)In\left(\frac{380K}{280K}\right)}\phantom{\rule{0ex}{0ex}}=449.8\mathrm{J}/\mathrm{kgK}\phantom{\rule{0ex}{0ex}}~450\mathrm{J}/\mathrm{kgK}$

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

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