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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

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 $3.5 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$.
Suppose you mix 4.0 mol of a monatomic gas at \(20.0^{\circ} \mathrm{C}\) and 3.0 mol of another monatomic gas at \(30.0^{\circ} \mathrm{C} .\) If the mixture is allowed to reach equilibrium, what is the final temperature of the mixture? [Hint: Use energy conservation.]
A reversible engine with an efficiency of \(30.0 \%\) has $T_{\mathrm{C}}=310.0 \mathrm{K} .\( (a) What is \)T_{\mathrm{H}} ?$ (b) How much heat is exhausted for every \(0.100 \mathrm{kJ}\) of work done?
What is the efficiency of an electric generator that produces $1.17 \mathrm{kW}\(.h per \)\mathrm{kg}$ of coal burned? The heat of combustion of coal is \(6.71 \times 10^{6} \mathrm{J} / \mathrm{kg}\).
A reversible refrigerator has a coefficient of performance of \(3.0 .\) How much work must be done to freeze \(1.0 \mathrm{kg}\) of liquid water initially at \(0^{\circ} \mathrm{C} ?\)
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