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Chapter 19: Problem 57

An object's heat capacity is inversely proportional to its absolute temperature: \(C=C_{0}\left(T_{0} / T\right),\) where \(C_{0}\) and \(T_{0}\) are constants. Find the entropy change when the object is heated from \(T_{0}\) to \(T_{1}\).

Short Answer

Expert verified
\(\Delta S = C_0 (-1 + \frac{T_0}{T_1})\) is the entropy change when the object is heated from \(T_0\) to \(T_1\).

Step by step solution

01

Recognize the given formula

The formula for heat capacity is given by \(C=C_{0}\left(T_{0} / T\right)\). This is provided in the question. Heat capacity \(C\) is inversely proportional to the absolute temperature \(T\), and \(C_{0}\) and \(T_{0}\) are constants.
02

Relationship between heat capacity and entropy

The change in entropy (∆S) can be found using the formula \(\Delta S = \int_{T_0}^{T_1} \frac{C}{T} dT\). This formula derives from the second law of thermodynamics which states that change in entropy is the integral of the heat capacity divided by the temperature, over the change in temperature.
03

Substitute and Integrate

Substitute \(C=C_{0}\left(T_{0} / T\right)\) into the entropy change formula to get: \(\Delta S = \int_{T_0}^{T_1} \frac{C_0 (T_0 / T)}{T} dT = C_0 T_0 \int_{T_0}^{T_1} \frac{1}{T^2} dT\). This integral can be solved, providing \(-1/T\) evaluated from \(T_0\) to \(T_1\), which simplifies to \(-1/T_1 + 1/T_0\).
04

Calculation and simplification

Next, we calculate the entropy change: \(\Delta S = C_0 T_0 (-\frac{1}{T_1} + \frac{1}{T_0}) = C_0 (-1 + \frac{T_0}{T_1})\).

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Most popular questions from this chapter

Use appropriate energy-flow diagrams to analyze the situation in Got It? \(19.2 ;\) that is, show that using a refrigerator to cool the lowtemperature reservoir can't increase the overall efficiency of a Carnot engine when the work input to the refrigerator is included.

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