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Chapter 1: Problem 20

A 60-gallon water heater is initially filled with water at \(50^{\circ} \mathrm{F}\). Determine how much energy (in Btu) needs to be transferred to the water to raise its temperature to \(120^{\circ} \mathrm{F}\). Evaluate the water properties at an average water temperature of \(85^{\circ} \mathrm{F}\).

Short Answer

Expert verified
Answer: The energy required is 34,951 Btu.

Step by step solution

01

Convert gallons to mass

First, we need to convert the volume of water (60 gallons) to its mass. We will use the density of water to do this. At \(85^{\circ} \mathrm{F}\), the density of water is approximately 62.3 lb/ft\(^{3}\) (according to the Engineering Toolbox). One gallon is equal to 0.133681 ft\(^{3}\), so we can use this conversion factor to find the mass of water. \(m = \text{Volume} \times \text{Density} = 60 ~\text{gallons} \times 0.133681 ~\text{ft}^{3}/\text{gallon} \times 62.3 ~\text{lb/ft}^{3} = 499.3 ~\text{lb}\)
02

Find the specific heat capacity of water

Next, we need to find the specific heat capacity of water at the average temperature of \(85^{\circ} \mathrm{F}\). In British Thermal Units (Btu), the specific heat capacity of water is approximately 1 Btu/lb\(\cdot^{\circ} \mathrm{F}\).
03

Calculate the change in temperature

Now, we need to find the change in temperature, which is the difference between the final temperature and the initial temperature. \(\Delta T = T_{\text{final}} - T_{\text{initial}} = 120^{\circ} \mathrm{F} - 50^{\circ} \mathrm{F} = 70^{\circ} \mathrm{F}\)
04

Calculate the energy transferred

Finally, use the formula \(Q = mc\Delta T\) to find the energy transferred to the water. \(Q = (499.3 ~\text{lb}) \times (1 ~\text{Btu/lb} \cdot^{\circ} \mathrm{F}) \times (70^{\circ} \mathrm{F}) = 34,951 ~\text{Btu}\) The energy required to heat the water to \(120^{\circ} \mathrm{F}\) is 34,951 Btu.

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Most popular questions from this chapter

A 5-cm-external-diameter, 10-m-long hot-water pipe at \(80^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(5^{\circ} \mathrm{C}\) by natural convection with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat loss from the pipe by natural convection. Answer: \(2945 \mathrm{~W}\)

What is stratification? Is it likely to occur at places with low or high ceilings? How does it cause thermal discomfort for a room's occupants? How can stratification be prevented?

A cold bottled drink ( \(\left.m=2.5 \mathrm{~kg}, c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(5^{\circ} \mathrm{C}\) is left on a table in a room. The average temperature of the drink is observed to rise to \(15^{\circ} \mathrm{C}\) in 30 minutes. The average rate of heat transfer to the drink is (a) \(23 \mathrm{~W}\) (b) \(29 \mathrm{~W}\) (c) \(58 \mathrm{~W}\) (d) \(88 \mathrm{~W}\) (e) \(122 \mathrm{~W}\)

A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is (a) \(2 \mathrm{~kJ}\) (b) \(100 \mathrm{~kJ}\) (c) \(6000 \mathrm{~kJ}\) (d) \(7200 \mathrm{~kJ}\) (e) \(12,000 \mathrm{~kJ}\)

Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a \(60-\mathrm{W}\) lightbulb. You will need a thermometer, which can be purchased in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light and is transmitted by the glass, calculate the heat transfer coefficient from Newton's law of cooling.
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