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Chapter 15: Problem 57

An engincer designs a ship that gets its power in the following way: The engine draws in warm water from the ocean, and after extracting some of the water's internal energy, returns the water to the ocean at a temperature \(14.5^{\circ} \mathrm{C}\) lower than the ocean temperature. If the ocean is at a uniform temperature of \(17^{\circ} \mathrm{C},\) is this an efficient engine? Will the engineer's design work?

Short Answer

Expert verified
Explain your reasoning. The design of the ship's engine is not efficient and is unlikely to work well because the maximum possible efficiency, calculated using the Carnot Efficiency formula, is only 5%. This low efficiency is not sufficient for practical applications, indicating that the engineer should explore other designs and concepts for the ship's engine.

Step by step solution

01

State the Given Information

The engine draws in warm water from the ocean at a temperature of \(17^{\circ}\mathrm{C}\). The engine returns the water at a temperature of \(14.5^{\circ}\mathrm{C}\) lower than the ocean temperature.
02

Convert Temperatures to Kelvin

We'll need to convert the given temperatures in Celsius to Kelvin, as it's necessary for thermodynamic calculations. The conversion formula is: \(K = C + 273.15.\) Ocean temperature (in Kelvin): \(17^{\circ}\mathrm{C} + 273.15 = 290.15 K\) Returned water temperature (in Kelvin): \((17 - 14.5)^{\circ}\mathrm{C} + 273.15 = 275.65 K\)
03

Calculate Efficiency using Carnot Efficiency

Since it is a heat engine, we will calculate the maximum possible efficiency using the Carnot Efficiency formula: Carnot Efficiency = \(1 - \frac{T_C}{T_H}\) Where \(T_H\) is the temperature of the hot reservoir (ocean) and \(T_C\) is the temperature of the cold reservoir (returned water). Carnot Efficiency = \(1 - \frac{275.65}{290.15} = 1 - 0.9500 = 0.0500\) The maximum possible efficiency of this engine is \(5\%\).
04

Evaluate the Efficiency and Design

A \(5\%\) efficiency is very low for an engine. Therefore, from the perspective of efficiency, this engine design is not efficient and would not work well. In conclusion, the engineer's design of a ship's engine that extracts energy from the temperature difference in water is not efficient and most likely will not work well. The low efficiency of \(5\%\) is not sufficient for practical applications, and the engineer should explore other designs and concepts for the ship's engine.

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Most popular questions from this chapter

An ice cube at \(0.0^{\circ} \mathrm{C}\) is slowly melting. What is the change in the ice cube's entropy for each \(1.00 \mathrm{g}\) of ice that melts?
The internal energy of a system increases by 400 J while \(500 \mathrm{J}\) of work are performed on it. What was the heat flow into or out of the system?
A \(0.50-\mathrm{kg}\) block of iron $[c=0.44 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]\( at \)20.0^{\circ} \mathrm{C}$ is in contact with a \(0.50-\mathrm{kg}\) block of aluminum \([c=\) $0.900 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]$ at a temperature of \(20.0^{\circ} \mathrm{C} .\) The system is completely isolated from the rest of the universe. Suppose heat flows from the iron into the aluminum until the temperature of the aluminum is \(22.0^{\circ} \mathrm{C}\) (a) From the first law, calculate the final temperature of the iron. (b) Estimate the entropy change of the system. (c) Explain how the result of part (b) shows that this process is impossible. [Hint: since the system is isolated, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$
A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)
A coal-fired electrical generating station can use a higher \(T_{\mathrm{H}}\) than a nuclear plant; for safety reasons the core of a nuclear reactor is not allowed to get as hot as coal. Suppose that $T_{\mathrm{H}}=727^{\circ} \mathrm{C}\( for a coal station but \)T_{\mathrm{H}}=527^{\circ} \mathrm{C}$ for a nuclear station. Both power plants exhaust waste heat into a lake at \(T_{\mathrm{C}}=27^{\circ} \mathrm{C} .\) How much waste heat does each plant exhaust into the lake to produce \(1.00 \mathrm{MJ}\) of electricity? Assume both operate as reversible engines.
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