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Chapter 15: Problem 35

An electric power station generates steam at \(500.0^{\circ} \mathrm{C}\) and condenses it with river water at \(27^{\circ} \mathrm{C} .\) By how much would its theoretical maximum efficiency decrease if it had to switch to cooling towers that condense the steam at \(47^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The decrease in the theoretical maximum efficiency is approximately 0.02533 or 2.533%.

Step by step solution

01

Convert temperatures from Celsius to Kelvin

To convert the given temperatures from Celsius to Kelvin, we just need to add 273.15 to each value. Converting the temperatures, we have: $T_1 = 500.0^{\circ} \mathrm{C} + 273.15 = 773.15 \mathrm{K} \newline T_2 = 27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K} \newline T_3 = 47^{\circ} \mathrm{C} + 273.15 = 320.15 \mathrm{K}$
02

Calculate initial maximum efficiency

Now that we have the temperatures in Kelvin, we can plug them into the formula for maximum efficiency. The initial efficiency is given by: Initial efficiency \(= 1 - \frac{T_2}{T_1} = 1 - \frac{300.15}{773.15} \approx 0.61153\)
03

Calculate new maximum efficiency with cooling towers

Similarly, we can calculate the new maximum efficiency with cooling towers using the formula: New efficiency \(= 1 - \frac{T_3}{T_1} = 1 - \frac{320.15}{773.15} \approx 0.5862\)
04

Calculate the decrease in efficiency

To find the decrease in efficiency, subtract the new efficiency value from the initial efficiency value: Decrease in efficiency \(= 0.61153 - 0.5862 \approx 0.02533\) So, the theoretical maximum efficiency of the electric power station would decrease by approximately 0.02533 or 2.533% if it had to switch to cooling towers that condense the steam at \(47^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

Consider a heat engine that is not reversible. The engine uses $1.000 \mathrm{mol}\( of a diatomic ideal gas. In the first step \)(\mathrm{A})$ there is a constant temperature expansion while in contact with a warm reservoir at \(373 \mathrm{K}\) from \(P_{1}=1.55 \times 10^{5} \mathrm{Pa}\) and $V_{1}=2.00 \times 10^{-2} \mathrm{m}^{3}$ to \(P_{2}=1.24 \times 10^{5} \mathrm{Pa}\) and $V_{2}=2.50 \times 10^{-2} \mathrm{m}^{3} .$ Then (B) a heat reservoir at the cooler temperature of \(273 \mathrm{K}\) is used to cool the gas at constant volume to \(273 \mathrm{K}\) from \(P_{2}\) to $P_{3}=0.91 \times 10^{5} \mathrm{Pa} .$ This is followed by (C) a constant temperature compression while still in contact with the cold reservoir at \(273 \mathrm{K}\) from \(P_{3}, V_{2}\) to \(P_{4}=1.01 \times 10^{5} \mathrm{Pa}, V_{1} .\) The final step (D) is heating the gas at constant volume from \(273 \mathrm{K}\) to \(373 \mathrm{K}\) by being in contact with the warm reservoir again, to return from \(P_{4}, V_{1}\) to \(P_{1}, V_{1} .\) Find the change in entropy of the cold reservoir in step \(\mathrm{B}\). Remember that the gas is always in contact with the cold reservoir. (b) What is the change in entropy of the hot reservoir in step D? (c) Using this information, find the change in entropy of the total system of gas plus reservoirs during the whole cycle.
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, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$
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?
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. \(]\)
Estimate the entropy change of \(850 \mathrm{g}\) of water when it is heated from \(20.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C} .\) [Hint: Assume that the heat flows into the water at an average temperature. \(]\)
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