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Chapter 15: Problem 24

An air conditioner whose coefficient of performance is 2.00 removes $1.73 \times 10^{8} \mathrm{J}$ of heat from a room per day. How much does it cost to run the air conditioning unit per day if electricity costs \(\$ 0.10\) per kilowatt-hour? (Note that 1 kilowatt-hour \(=3.6 \times 10^{6} \mathrm{J} .\) )

Short Answer

Expert verified
Answer: To find the daily cost of running the air conditioning unit, calculate the final value using the formula: Daily cost = \(( (1.73 \times 10^{8} \mathrm{J}) / 2.00) \cdot (1 \mathrm{kWh}) / (3.6 \times 10^{6} \mathrm{J}) \times \$0.10/\mathrm{kWh}\) .

Step by step solution

01

Calculate work done by the air conditioner per day (W)

We will use the formula for the coefficient of performance (COP) of the air conditioner to find the work done: COP = (Heat removed per day) / (Work done per day) Work done per day (W) = (Heat removed per day) / COP W = \((1.73 \times 10^{8} \mathrm{J}) / 2.00\)
02

Convert work done per day (W) to kilowatt-hours (kWh)

Now, convert the work done per day (W) from Joules to kilowatt-hours using the given conversion: 1 kilowatt-hour = \(3.6 \times 10^{6} \mathrm{J}\) W (kWh) = \((1.73 \times 10^{8} \mathrm{J}) / 2.00 \cdot (1 \mathrm{kWh}) / (3.6 \times 10^{6} \mathrm{J})\)
03

Calculate the daily cost of running the air conditioner

Finally, calculate the daily cost by multiplying the kilowatt-hours (kWh) with the electricity cost per kilowatt-hour: Daily cost = W (kWh) × cost per kilowatt-hour Daily cost = \(( (1.73 \times 10^{8} \mathrm{J}) / 2.00) \cdot (1 \mathrm{kWh}) / (3.6 \times 10^{6} \mathrm{J}) \times \$0.10/\mathrm{kWh}\) Calculate the final value to get the cost of running the air conditioning unit per day.

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Most popular questions from this chapter

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, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$
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.
(a) Calculate the efficiency of a reversible engine that operates between the temperatures \(600.0^{\circ} \mathrm{C}\) and \(300.0^{\circ} \mathrm{C} .\) (b) If the engine absorbs \(420.0 \mathrm{kJ}\) of heat from the hot reservoir, how much does it exhaust to the cold reservoir?
A town is considering using its lake as a source of power. The average temperature difference from the top to the bottom is \(15^{\circ} \mathrm{C},\) and the average surface temperature is \(22^{\circ} \mathrm{C} .\) (a) Assuming that the town can set up a reversible engine using the surface and bottom of the lake as heat reservoirs, what would be its efficiency? (b) If the town needs about \(1.0 \times 10^{8} \mathrm{W}\) of power to be supplied by the lake, how many \(\mathrm{m}^{3}\) of water does the heat engine use per second? (c) The surface area of the lake is $8.0 \times 10^{7} \mathrm{m}^{2}$ and the average incident intensity (over \(24 \mathrm{h})\) of the sunlight is \(200 \mathrm{W} / \mathrm{m}^{2} .\) Can the lake supply enough heat to meet the town's energy needs with this method?
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?
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