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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Transfer
In thermodynamics, energy transfer refers to the process of moving energy from one system to another. When heating water, energy is added to increase the water's temperature. This energy transfer is typically measured in British Thermal Units (Btu), which quantify the amount of heat needed to raise the temperature.

In the context of our problem, energy is transferred to the water heater to increase its temperature from 50掳F to 120掳F. This requires us to calculate the total amount of energy needed, which involves understanding concepts like specific heat capacity and temperature difference. In this exercise, the energy required is shown to be 34,951 Btu. This is the amount of energy that needs to be provided to the water to achieve the desired temperature rise, demonstrating the importance of understanding energy transfer in processes like heating.
Specific Heat Capacity
Specific heat capacity is a crucial concept in thermodynamics, and it represents the amount of energy required to raise the temperature of one pound of a substance by one degree Fahrenheit. For water, this value is particularly important because of its use in heating applications.

Water has a specific heat capacity of about 1 Btu/lb路掳F at the average temperature considered in this exercise (85掳F). This relatively high specific heat capacity means that water can store and transfer a large amount of energy compared to many other substances. It's why water is often used in heating and cooling systems.

Understanding the specific heat capacity allows us to calculate how much energy is required to increase the water's temperature, using the formula:
  • \(Q = mc\Delta T\)

where \(m\) is the mass of the water, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. This concept helps us predict the energy needs of systems involving thermal energy conversion.
Thermal Properties of Water
The thermal properties of water make it unique and invaluable in thermal applications. One of the key properties is its high specific heat capacity, meaning it can absorb a lot of heat without a significant rise in temperature. This characteristic of water is beneficial for regulating temperatures in various settings.

In the exercise, we also consider the density of water at 85掳F, which is approximately 62.3 lb/ft鲁. This density allows us to convert the volume of water into mass, which is crucial for using the formula to calculate energy transfer. Without knowing the mass, it would be impossible to determine how much energy is needed to heat the water.

Another important aspect of water is its relatively consistent properties over a range of temperatures, which simplifies calculations in practical applications. Understanding these thermal properties helps us efficiently design and operate systems that rely on water's ability to store and transfer heat.

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Most popular questions from this chapter

The inner and outer surfaces of a 4-m \(\times 7-\mathrm{m}\) brick wall of thickness \(30 \mathrm{~cm}\) and thermal conductivity \(0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) are maintained at temperatures of \(26^{\circ} \mathrm{C}\) and \(8^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer through the wall, in W.

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{~kg} / \mathrm{s}\) and is heated to \(60^{\circ} \mathrm{C}\). The rate of heat transfer to the water is (a) \(20 \mathrm{~kW}\) (b) \(42 \mathrm{~kW}\) (c) \(84 \mathrm{~kW}\) (d) \(126 \mathrm{~kW}\) (e) \(334 \mathrm{~kW}\)

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is \(10^{\circ} \mathrm{C}\), while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat loss by radiation.

Write an essay on how microwave ovens work, and explain how they cook much faster than conventional ovens. Discuss whether conventional electric or microwave ovens consume more electricity for the same task.
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