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Chapter 6: Problem 9

How much will it cost per day to keep a house at \(20^{\circ} \mathrm{C}\) inside when the external temperature is steady at \(-5^{\circ} \mathrm{C}\) using direct electric heating \(^{65}\) if the house is rated at \(150 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) and electricity costs \(\$ 0.15 / \mathrm{kWh} ?\)

Short Answer

Expert verified
Answer: The daily cost to keep the house at this temperature is $20.77.

Step by step solution

01

Find the temperature difference

First, we will find out the temperature difference between the inside of the house and the outside. Temperature difference = Inside temperature - Outside temperature Temperature difference = \(20^{\circ}\mathrm{C} - (-5^{\circ}\mathrm{C}) = 25^{\circ}\mathrm{C}\)
02

Calculate the energy needed to maintain the temperature difference

We are given that the house is rated at \(150\mathrm{~W}/{ }^{\circ}\mathrm{C}\). This means that to maintain a \(1^{\circ}\mathrm{C}\) difference, we need 150 W. To find the energy needed to maintain the calculated temperature difference of 25 degrees, we will multiply the power needed for 1-degree difference by the temperature difference. Energy needed = (Power per \(1^{\circ}\mathrm{C}\)) × (Temperature difference) Energy needed = \(150\mathrm{~W}/{ }^{\circ}\mathrm{C} \times 25^{\circ}\mathrm{C}\) Energy needed = 3750 W
03

Account for the effectiveness of the heating system

We are given that the heater efficiency is 65%, meaning only 65% of the energy consumed is used for heating. First, we'll find the equivalent energy for 100% efficiency which can be done by dividing the actual energy needed by the efficiency percentage. Equivalent Energy = \(\frac{Energy\:needed}{Efficiency}\) Equivalent Energy = \(\frac{3750\:W}{0.65}\) Equivalent Energy = 5769.23 W
04

Convert the energy consumption to kilowatts-hours per day

Convert the Watts to kilowatts and then multiply by 24 to get the daily energy consumption in kilowatt-hours (kWh). Daily energy consumption = \(\frac{Energy\:needed}{1000} \times 24 \mathrm{hours}\) Daily energy consumption = \(\frac{5769.23\:W}{1000} \times 24 \mathrm{hours}\) Daily energy consumption = 138.4615 kWh
05

Calculate the daily cost

To find the daily cost, we will multiply the daily energy consumption in kWh by the cost of electricity per kWh. Daily cost = (Daily energy consumption) × (Cost of electricity per kWh) Daily cost = 138.4615 kWh × \(0.15 / \mathrm{kWh}\) Daily cost = $20.77 So, the daily cost to keep the house at \(20^{\circ}\mathrm{C}\) when the external temperature is \(-5^{\circ}\mathrm{C}\) using direct electric heating is $20.77.

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Most popular questions from this chapter

A heat engine pulls \(100 \mathrm{~J}\) out of a hot bath at \(800 \mathrm{~K}\), and transfers \(80 \mathrm{~J}\) of heat into the cold bath at \(300 \mathrm{~K}\). What efficiency does this heat engine achieve in producing useful work, and how does it compare to the theoretical maximum?

In a house achieving a heat loss rate of \(200 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) equipped a \(5,000 \mathrm{~W}\) heater, what will the internal temperature be if the outside temperature is \(-10^{\circ} \mathrm{C}\) and the heater is running \(100 \%\) of the time?

What would the maximum thermodynamic efficiency be of some heat engine operating between your skin temperature and the ambient environment \(20^{\circ} \mathrm{C}\) cooler than your skin?

Provide at least one example not listed in the text in which heat flows into some other form of energy. \(^{66}\) In the text, we mentioned hot air over a car, wind, internal combustion, and a steam turbine plant.

Changing from direct electrical heating to a heat pump operating with a COP of 3 means spending one-third the energy for a certain thermal benefit. If a house averages \(30 \mathrm{kWh} /\) day in heating cost through the year using direct electrical heating at a cost of \(\$ 0.15 / \mathrm{kWh}\), how long will it take to recuperate a \(\$ 5,000\) installation cost of a new heat pump?
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