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Chapter 15: Problem 21

An engine releases \(0.450 \mathrm{kJ}\) of heat for every \(0.100 \mathrm{kJ}\) of work it does. What is the efficiency of the engine?

Short Answer

Expert verified
Answer: The efficiency of the engine is 18.18%.

Step by step solution

01

Calculate the Total Energy Input

Given that the engine releases 0.450 kJ of heat for every 0.100 kJ of work it does, we can determine the total energy input by summing the energy of work and heat. Total Energy Input = Work Output + Heat Released = 0.100 kJ + 0.450 kJ = 0.550 kJ.
02

Calculate Efficiency

Now, we can calculate the efficiency of the engine using the formula: Efficiency = (Work Output / Total Energy Input) * 100. Efficiency = (0.100 kJ / 0.550 kJ) * 100 = 18.18%. Thus, the efficiency of the engine is 18.18%.

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Most popular questions from this chapter

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.
List these in order of increasing entropy: (a) 1 mol of water at $20^{\circ} \mathrm{C}\( and 1 mol of ethanol at \)20^{\circ} \mathrm{C}$ in separate containers; (b) a mixture of 1 mol of water at \(20^{\circ} \mathrm{C}\) and 1 mol of ethanol at \(20^{\circ} \mathrm{C} ;\) (c) 0.5 mol of water at \(20^{\circ} \mathrm{C}\) and 0.5 mol of ethanol at \(20^{\circ} \mathrm{C}\) in separate containers; (d) a mixture of 1 mol of water at $30^{\circ} \mathrm{C}\( and 1 mol of ethanol at \)30^{\circ} \mathrm{C}$.
In a certain steam engine, the boiler temperature is \(127^{\circ} \mathrm{C}\) and the cold reservoir temperature is \(27^{\circ} \mathrm{C} .\) While this engine does \(8.34 \mathrm{kJ}\) of work, what minimum amount of heat must be discharged into the cold reservoir?
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 $3.5 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$.
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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