Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 15, Problem 74P

(a)

To determine

The efficiency of reversible engine.

(a)

Expert Solution
Check Mark

Answer to Problem 74P

Efficiency is 0.051.

Explanation of Solution

Average temperature difference between top and bottom is 15°C and the average surface temperature is 22°C.

Write the equation to calculate the efficiency.

ε=1(TsurTdiffTsur)

Here, ε is the efficiency, Tsur is the average surface temperature and Tdiff is the average temperature difference between top and bottom.

Conclusion:

Substitute 22°C for Tsur and 15°C or Tdiff in the above equation to find ε.

ε=1(22°C15°C22°C15°C)(273.15K+22K15K)(22°C22°C)(273.15K+22K)=10.949=0.051

Therefore, the efficiency is 0.051.

(b)

To determine

The cubic meters of water used by heat engine per second.

(b)

Expert Solution
Check Mark

Answer to Problem 74P

Water used by heat engine is 31m3.

Explanation of Solution

Average temperature difference between top and bottom is 15°C and the average surface temperature is 22°C, power supplied by the lake is 1.0×108W.

Write the equation to find the power supplied to the town.

P=εΔQΔt

Here, P is the power supplied to the town, ΔQ is the heat energy supplied, and Δt is the time duration.

Write the equation to find ΔQ.

ΔQ=(ρV)cΔT

Here, ρ is the density of water, V is the volume of water, c is the specific heat capacity of water, and ΔT is the temperature change.

Rewrite the equation for P by substituting the above relation for ΔQ.

P=ε(ρV)cΔTΔtV=PΔteρcΔT

Conclusion:

Substitute 1.0×108W for P, 1.0s for Δt, 0.051 for ε, 1.00×103kg/m3 for ρ, 4186J/kgK for c, and 15K for ΔT in the above equation to find V.

V=(1.0×108W)(1.0s)(0.051)(1.00×103kg/m3)(4186J/kgK)(15K)=31m3

Therefore, the water used by heat engine is 31m3.

(c)

To determine

Check whether the lake can supply enough heat energy to meet the town’s energy need.

(c)

Expert Solution
Check Mark

Answer to Problem 74P

Yes. Lake can supply enough heat energy to meet the town’s energy need.

Explanation of Solution

Average temperature difference between top and bottom is 15°C and the average surface temperature is 22°C, power supplied by the lake is 1.0×108W, surface area of lake is 8.0×107m2, and the average incident intensity is 200W/m2.

Write the equation to find the power delivered by sun.

Psun=IA

Here, Psun is the average power of sunlight, I is the intensity, and A is the area of cross section.

Write the equation to calculate the power of engine.

Pengine=Ptownε

Here, Pengine is the power of engine, and Ptown is the power required by the town.

Conclusion:

Substitute 200W/m2 for I and 8.0×107m2 for A in the equation for Psun.

Psun=(200W/m2)(8.0×107m2)=1.6×1010W

Substitute 1.0×108W for Ptown and 0.051 for ε in the equation for Pengine.

Pengine=1.0×108W0.051=2.0×109W

It can be seen that Psun>Pengine.This means that the lake can supply enough heat energy to meet the town’s energy need.

Therefore, Yes. Lake can supply enough heat energy to meet the town’s energy need.

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Chapter 15 Solutions

Physics

Ch. 15.8 - Prob. 15.8CPCh. 15.8 - Prob. 15.9PPCh. 15 - Prob. 1CQCh. 15 - Prob. 2CQCh. 15 - Prob. 3CQCh. 15 - Prob. 4CQCh. 15 - Prob. 5CQCh. 15 - Prob. 6CQCh. 15 - Prob. 7CQCh. 15 - Prob. 8CQCh. 15 - Prob. 9CQCh. 15 - Prob. 10CQCh. 15 - 11. A warm pitcher of lemonade is put into an ice...Ch. 15 - Prob. 12CQCh. 15 - Prob. 13CQCh. 15 - Prob. 14CQCh. 15 - Prob. 1MCQCh. 15 - Prob. 2MCQCh. 15 - Prob. 3MCQCh. 15 - Prob. 4MCQCh. 15 - Prob. 5MCQCh. 15 - Prob. 6MCQCh. 15 - Prob. 7MCQCh. 15 - Prob. 8MCQCh. 15 - Prob. 9MCQCh. 15 - Prob. 10MCQCh. 15 - Prob. 11MCQCh. 15 - Prob. 12MCQCh. 15 - Prob. 13MCQCh. 15 - Prob. 1PCh. 15 - Prob. 2PCh. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - Prob. 5PCh. 15 - Prob. 6PCh. 15 - Prob. 7PCh. 15 - Prob. 8PCh. 15 - Prob. 9PCh. 15 - Prob. 10PCh. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Prob. 13PCh. 15 - Prob. 14PCh. 15 - Prob. 15PCh. 15 - Prob. 16PCh. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - Prob. 19PCh. 15 - Prob. 20PCh. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 23PCh. 15 - Prob. 24PCh. 15 - 25. What is the efficiency of an electric...Ch. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - Prob. 28PCh. 15 - Prob. 29PCh. 15 - Prob. 30PCh. 15 - Prob. 31PCh. 15 - Prob. 32PCh. 15 - Prob. 33PCh. 15 - Prob. 34PCh. 15 - Prob. 35PCh. 15 - Prob. 36PCh. 15 - Prob. 37PCh. 15 - Prob. 38PCh. 15 - Prob. 39PCh. 15 - Prob. 40PCh. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - Prob. 43PCh. 15 - Prob. 44PCh. 15 - Prob. 45PCh. 15 - Prob. 46PCh. 15 - Prob. 47PCh. 15 - Prob. 48PCh. 15 - Prob. 49PCh. 15 - Prob. 50PCh. 15 - Prob. 51PCh. 15 - Prob. 52PCh. 15 - Prob. 53PCh. 15 - Prob. 55PCh. 15 - Prob. 54PCh. 15 - Prob. 56PCh. 15 - Prob. 57PCh. 15 - Prob. 58PCh. 15 - Prob. 59PCh. 15 - Prob. 60PCh. 15 - Prob. 61PCh. 15 - Prob. 62PCh. 15 - Prob. 63PCh. 15 - Prob. 64PCh. 15 - Prob. 65PCh. 15 - Prob. 66PCh. 15 - Prob. 67PCh. 15 - Prob. 68PCh. 15 - Prob. 69PCh. 15 - Prob. 70PCh. 15 - Prob. 71PCh. 15 - Prob. 73PCh. 15 - Prob. 72PCh. 15 - Prob. 74PCh. 15 - Prob. 75PCh. 15 - Prob. 76PCh. 15 - Prob. 77PCh. 15 - Prob. 78PCh. 15 - Prob. 79PCh. 15 - Prob. 80PCh. 15 - Prob. 81PCh. 15 - Prob. 82PCh. 15 - Prob. 83PCh. 15 - Prob. 84PCh. 15 - Prob. 85PCh. 15 - Prob. 86PCh. 15 - Prob. 87PCh. 15 - Prob. 88PCh. 15 - Prob. 89PCh. 15 - Prob. 90PCh. 15 - Prob. 91PCh. 15 - Prob. 92PCh. 15 - Prob. 93PCh. 15 - Prob. 94PCh. 15 - Prob. 95PCh. 15 - Prob. 96PCh. 15 - Prob. 97PCh. 15 - Prob. 98PCh. 15 - Prob. 99PCh. 15 - Prob. 100P
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