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Chapter 19: Problem 16

A Carnot engine absorbs \(900 \mathrm{J}\) of heat each cycle and provides \(350 \mathrm{J}\) of work. (a) What's its efficiency? (b) How much heat is rejected each cycle? (c) If the engine rejects heat at \(10^{\circ} \mathrm{C},\) what's its maximum temperature?

Short Answer

Expert verified
The Carnot engine has an efficiency of 38.88%, rejects 550 J of heat each cycle, and its maximum temperature is 463.69 K.

Step by step solution

01

Calculate the Efficiency

The efficiency of any heat engine is defined as the ratio of the work it does to the heat absorbed during one cycle. In mathematical terms, it is \[ \text{Efficiency} = \frac{\text{Work}}{\text{Heat absorbed}} \]. Therefore, by inserting the given values, we get \[ \text{Efficiency} = \frac{350 \, \text{J}}{900 \, \text{J}} = 0.3888 = 38.88 \% \]. This is the engine's efficiency.
02

Calculate the Heat Rejected

The heat rejected by the engine can be calculated as the difference between the heat absorbed and the work done. So, \[ \text{Heat rejected} = \text{Heat absorbed} - \text{Work} = 900 \, \text{J} - 350 \, \text{J} = 550 \, \text{J} \] . This is the heat rejected by the engine in each cycle.
03

Calculate the Maximum Temperature

The efficiency of a Carnot engine can also be expressed in terms of the temperatures of the heat reservoirs it operates between, according to the formula \[ \text{Efficiency} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \] where the temperatures are in Kelvin. So we can rearrange the formula to solve for the hot temperature: \[ T_{\text{hot}}=\frac{T_{\text{cold}}}{1-\text{Efficiency}} \]. The cold temperature is given as \( 10^{\circ} \mathrm{C} \) , which corresponds to \( 283.15 \, \text{ K} \) . Inserting the values, we get \[ T_{\text{hot}}=\frac{283.15 \, \text{ K}}{1-0.3888}= 463.69 \, \text{ K} \] . That's the maximum temperature of the carnot engine.

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