How much energy becomes unavailable for work in an isothermal process at \(440 \mathrm{K},\) if the entropy increase is \(25 \mathrm{J} / \mathrm{K} ?\)

Short Answer

Expert verified
The energy that becomes unavailable for work in this isothermal process is 11,000 Joules.

Step by step solution

01

Understand the Problem

The problem asks for the energy that becomes unavailable for work. Given are the temperature and the increase in entropy during the process. The formula to follow is \( EnergyUnavailability = T * \Delta S \), where \( T \) is the temperature and \(\Delta S \) is the change in entropy.
02

Substitution

Substitute the given values into the formula: \( EnergyUnavailability = 440 K * 25 J/K \)
03

Calculate

Solve the equation for \( EnergyUnavailability\), which gives \( EnergyUnavailability = 11,000 J \)

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Most popular questions from this chapter

A reversible engine contains 0.20 mol of ideal monatomic gas, initially at \(600 \mathrm{K}\) and confined to \(2.0 \mathrm{L} .\) The gas undergoes the following cycle: Isothermal expansion to \(4.0 \mathrm{L}\) \(\cdot\) Isovolumic cooling to \(300 \mathrm{K}\) Isothermal compression to \(2.0 \mathrm{L}\) \(\cdot\) Isovolumic heating to \(600 \mathrm{K}\) (a) Calculate the net heat added during the cycle and the net work done. (b) Determine the engine's efficiency, defined as the ratio of the work done to the heat absorbed during the cycle.

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