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Chapter 24: Problem 28

An inventor claims to have created a heat pump that draws heat from a lake at \(3.0^{\circ} \mathrm{C}\) and delivers heat at a rate of \(20 \mathrm{~kW}\) to a building at \(35^{\circ} \mathrm{C}\), while using only \(1.9 \mathrm{~kW}\) of electrical power. How would you judge the claim?

Short Answer

Expert verified
The claim is likely inaccurate, as the claimed efficiency of the heat pump is greater than the maximum efficiency allowed by the second law of thermodynamics (the Carnot efficiency).

Step by step solution

01

Convert temperatures to Kelvin

We first need to convert all temperatures to Kelvin. The scale of Kelvin starts at absolute zero and each degree Kelvin is exactly the size of a degree Celsius. The difference is that 0 Kelvin is equivalent to -273.15 Celsius. Thus, to convert a temperature in Celsius to Kelvin, add 273.15. So, \(T_{hot} = 35^{\circ} \mathrm{C} + 273.15 = 308.15 \mathrm{K}\) and \(T_{cold} = 3^{\circ} \mathrm{C} + 273.15 = 276.15 \mathrm{K}\)
02

Calculate the Carnot efficiency

The maximum possible efficiency for a heat pump (the Carnot efficiency) is \((T_{hot}-T_{cold})/T_{hot}\). Substituting the values that we got in the first step we get \((308.15 - 276.15) / 308.15 = 0.104\). So, the Carnot efficiency is \(0.104\), or \(10.4\%\)
03

Calculate the claimed efficiency of the heat pump

The claimed power output of the heat pump is \(20 \mathrm{~kW}\), and the electrical power input is \(1.9 \mathrm{~kW}\). Thus the claimed efficiency is \(20 / 1.9 = 10.53\)
04

Compare the Carnot and claimed efficiencies

The claimed efficiency, \(10.53\), is greater than the Carnot efficiency, \(10.4\%\). Since Carnot efficiency is the maximum theoretically possible, having a higher efficiency breaks the second law of thermodynamics. Therefore, the claim may not be accurate.

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Most popular questions from this chapter

In a Carnot cycle, the isothermal expansion of an ideal gas takes place at \(412 \mathrm{~K}\) and the isothermal compression at \(297 \mathrm{~K}\). During the expansion, \(2090 \mathrm{~J}\) of heat energy are transferred to the gas. Determine \((a)\) the work performed by the gas during the isothermal expansion, ( \(b\) ) the heat rejected from the gas during the isothermal compression, and \((c)\) the work done on the gas during the isothermal compression.

Calculate the efficiency of a fossil-fuel power plant that consumes 382 metric tons of coal each hour to produce useful work at the rate of \(755 \mathrm{MW}\). The heat of combustion of coal is \(28.0 \mathrm{MJ} / \mathrm{kg}\).

A mixture of \(1.78 \mathrm{~kg}\) of water and \(262 \mathrm{~g}\) of ice at \(0{ }^{\circ} \mathrm{C}\) is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass, is \(1: 1\) at \(0^{\circ} \mathrm{C} .(a)\) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

A car engine delivers \(8.18 \mathrm{~kJ}\) of work per cycle. (a) Before a tune-up, the efficiency is \(25.0 \% .\) Calculate, per cycle, the heat absorbed from the combustion of fuel and the heat exhausted to the atmosphere. (b) After a tune-up, the efficiency is \(31.0 \%\). What are the new values of the quantities calculated in \((a) ?\)

(a) A Carnot engine operates between a hot reservoir at \(322 \mathrm{~K}\) and a cold reservoir at \(258 \mathrm{~K}\). If it absorbs \(568 \mathrm{~J}\) of heat per cycle at the hot reservoir, how much work per cycle does it deliver? (b) If the same engine, working in reverse, functions as a refrigerator between the same two reservoirs, how much work per cycle must be supplied to transfer \(1230 \mathrm{~J}\) of heat from the cold reservoir?
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