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Fundamentals Of Physics
Found in: Page 604

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Short Answer

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 S of the block until thermal equilibrium is reached. The scale of the horizontal axis is set by Ta=280K and Tb=380K. What is the specific heat of the block?

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

See the step by step solution

Step by Step Solution

Step 1: 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, Ta=280K and Tb=380K

Step 2: 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 the specific heat of the block.


The entropy change of the gas, S=mclnTfTi …(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

Tf=380K,Ti=280K and S=J/K. )

c=SmlnTfTi =50 J/K0.364 kgIn380K280K =449.8J/kgK ~450J/kgK

Hence, the value of the specific heat of the block is 450J/kgK

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