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Chapter 6: Problem 13

What would the maximum thermodynamic efficiency be of some heat engine operating between your skin temperature and the ambient environment \(20^{\circ} \mathrm{C}\) cooler than your skin?

Short Answer

Expert verified
Answer: The formula for the maximum thermodynamic efficiency is Carnot efficiency = 1 - ((T_skin - 20) / T_skin), where T_skin is the skin temperature in Celsius.

Step by step solution

01

Convert temperatures to Kelvin

To convert temperatures to Kelvin, add 273.15 to the Celsius values. Let T_skin be the skin temperature in Celsius and T_ambient = T_skin - 20. Then, convert these temperatures to Kelvin using: T_skin_K = T_skin + 273.15 T_ambient_K = T_ambient + 273.15
02

Find the Carnot efficiency

The maximum thermodynamic efficiency of a heat engine is obtained from the Carnot efficiency formula: Carnot efficiency = 1 - (T_cold / T_hot) Where T_cold is the temperature of the cold reservoir (ambient environment) and T_hot is the temperature of the hot reservoir (skin). Plug in the T_ambient_K and T_skin_K values to find the maximum thermodynamic efficiency: Carnot efficiency = 1 - (T_ambient_K / T_skin_K)
03

Simplify the expression

Since both T_skin_K and T_ambient_K have the same constant added (273.15), we can rewrite the expression as Carnot efficiency = 1 - ((T_skin - 20) / T_skin) So, the maximum thermodynamic efficiency for this heat engine operating between skin temperature and ambient environment 20 degrees Celsius cooler than the skin is given by the above formula: Carnot efficiency = 1 - ((T_skin - 20) / T_skin)

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Most popular questions from this chapter

In a house achieving a heat loss rate of \(200 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) equipped a \(5,000 \mathrm{~W}\) heater, what will the internal temperature be if the outside temperature is \(-10^{\circ} \mathrm{C}\) and the heater is running \(100 \%\) of the time?

How many Joules does it take to heat your body up by \(1^{\circ} \mathrm{C}\) if your (water-dominated) mass has a specific heat capacity of \(3,500 \mathrm{~J} / \mathrm{kg} /{ }^{\circ} \mathrm{C} ?\)

Since the sun drives energy processes on Earth, we could explore the maximum possible thermodynamic efficiency of a process operating between the surface temperature of the sun \((5,800 \mathrm{~K})\) and Earth's surface temperature \((288 \mathrm{~K}) .\) What is this maximum efficiency? \(^{69}\)

If a can of soda \(\left(350 \mathrm{~mL} ;\right.\) treat as water) cools from \(20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\), how much energy is extracted, and how much is the entropy (in \(\mathrm{J} / \mathrm{K}\) ) in the can reduced using the average temperature and the relation that \(\Delta Q=T \Delta S ?\)

We can think of wind in the atmosphere as a giant heat engine \(^{67}\) operating between the \(288 \mathrm{~K}\) surface and the top of the troposphere \(^{68}\) at \(230 \mathrm{~K}\). What is the maximum efficiency this heat engine could achieve in converting solar heating into airflow?
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