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Chapter 15: Problem 28

Calculate the maximum possible efficiency of a heat engine that uses surface lake water at \(18.0^{\circ} \mathrm{C}\) as a source of heat and rejects waste heat to the water \(0.100 \mathrm{km}\) below the surface where the temperature is \(4.0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: The maximum possible efficiency of the heat engine is 4.81%.

Step by step solution

01

Convert temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, add 273.15 to the given temperature. Therefore, the surface lake water temperature is 18.0 + 273.15 = 291.15 K, and the deep water temperature is 4.0 + 273.15 = 277.15 K.
02

Calculate the Carnot efficiency

The Carnot efficiency is given by the formula: Efficiency = \(1 - \frac{T_{cold}}{T_{hot}}\) Where \(T_{cold}\) and \(T_{hot}\) are the temperatures of the cold and hot reservoirs, respectively. In this case, \(T_{hot} = 291.15 \mathrm{K}\) and \(T_{cold} = 277.15 \mathrm{K}\). Plugging these values into the formula, we get: Efficiency = \(1 - \frac{277.15}{291.15} = 1 - 0.9519 = 0.0481\)
03

Convert efficiency to percentage

To express the efficiency as a percentage, multiply the decimal value by 100: Maximum efficiency = \(0.0481 \times 100 = 4.81\%\) Thus, the maximum possible efficiency of the heat engine using the surface lake water at \(18.0^{\circ}\mathrm{C}\) as a source of heat and rejecting waste heat to the water \(0.100 \mathrm{km}\) below the surface where the temperature is \(4.0^{\circ}\mathrm{C}\) is 4.81%.

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Most popular questions from this chapter

A fish at a pressure of 1.1 atm has its swim bladder inflated to an initial volume of \(8.16 \mathrm{mL}\). If the fish starts swimming horizontally, its temperature increases from \(20.0^{\circ} \mathrm{C}\) to $22.0^{\circ} \mathrm{C}$ as a result of the exertion. (a) since the fish is still at the same pressure, how much work is done by the air in the swim bladder? [Hint: First find the new volume from the temperature change. \(]\) (b) How much heat is gained by the air in the swim bladder? Assume air to be a diatomic ideal gas. (c) If this quantity of heat is lost by the fish, by how much will its temperature decrease? The fish has a mass of \(5.00 \mathrm{g}\) and its specific heat is about $3.5 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$.
For a reversible engine, will you obtain a better efficiency by increasing the high-temperature reservoir by an amount \(\Delta T\) or decreasing the low- temperature reservoir by the same amount \(\Delta T ?\)
Show that in a reversible engine the amount of heat \(Q_{C}\) exhausted to the cold reservoir is related to the net work done \(W_{\text {net }}\) by $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} W_{\mathrm{net}} $$
How much heat does a heat pump with a coefficient of performance of 3.0 deliver when supplied with \(1.00 \mathrm{kJ}\) of electricity?
Within an insulated system, \(418.6 \mathrm{kJ}\) of heat is conducted through a copper rod from a hot reservoir at \(+200.0^{\circ} \mathrm{C}\) to a cold reservoir at \(+100.0^{\circ} \mathrm{C} .\) (The reservoirs are so big that this heat exchange does not change their temperatures appreciably.) What is the net change in entropy of the system, in \(\mathrm{kJ} / \mathrm{K} ?\)
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