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Chapter 1: Problem 74

(a) How much heat transfer is required to raise the temperature of a \(0.750-\mathrm{kg}\) aluminum pot containing \(2.50 \mathrm{kg}\) of water from \(30.0^{\circ} \mathrm{C}\) to the boiling point and then boil away \(0.750 \mathrm{kg}\) of water? (b) How long does this take if the rate of heat transfer is \(500 \mathrm{W} ?\)

Short Answer

Expert verified
The total heat transfer required to raise the temperature of a 0.750 kg aluminum pot containing 2.50 kg of water from 30掳C to the boiling point and then boil away 0.750 kg of water is 2464750 J. It will take 4929.5 seconds (approximately 82.15 minutes) for the heat transfer to occur at a rate of 500 W.

Step by step solution

01

: Calculate the Heat Needed to Raise the Temperature of the Pot and Water

: First, we need to find the heat necessary to raise the temperature of both the aluminum pot and the water inside to the boiling point. We do this using the heat transfer formula: \(Q = mc\Delta{T}\) Where Q is the heat transfer, m is the mass, c is the specific heat capacity, and \(\Delta{T}\) is the change in temperature. For the aluminum pot, the specific heat capacity is \(c_{aluminum} = 900 \frac{J}{kg \cdot 掳C}\), and for water, it is \(c_{water} = 4190 \frac{J}{kg \cdot 掳C}\). To find the heat needed for the pot: \(Q_{pot} = m_{pot}c_{aluminum} \Delta{T}\) To find the heat needed for the water: \(Q_{water} = m_{water}c_{water} \Delta{T}\)
02

: Calculate the Heat Needed to Boil Away Required Amount of Water

: Next, we need to determine the amount of heat needed to boil away 0.750 kg of water. This can be done using the latent heat of vaporization formula: \(Q = mL\) Where L is the latent heat of vaporization, which for water is \(L_{water} = 2.26 \times 10^6 \frac{J}{kg}\). To find the heat needed to boil away the water: \(Q_{boil} = m_{boil}L_{water}\)
03

: Calculate the Total Heat Transfer Required

: Now, we can find the total heat transfer required by summing the heat needed to raise the temperature of the pot and water, and the heat needed to boil away the required amount of water: \(Q_{total} = Q_{pot} + Q_{water} + Q_{boil}\)
04

: Calculate the Time Taken

: Finally, we need to calculate how long it takes to transfer the required amount of heat at the given rate of transfer, 500 W. Since power is defined as the rate of energy transfer per unit time, we can use the formula: \(t = \frac{Q_{total}}{P}\) Where t is the time and P is the power (rate of heat transfer). Now let's calculate the values. (a) Calculating heat transfer: The change in temperature is: \(\Delta{T} = 100^{\circ}C - 30^{\circ}C = 70^{\circ}C\) For the pot: \(Q_{pot} = (0.750 \,\mathrm{kg})(900 \,\frac{J}{kg \cdot 掳C})(70^{\circ}C) = 47250 \,\mathrm{J}\) For the water: \(Q_{water} = (2.50 \,\mathrm{kg})(4190 \,\frac{J}{kg \cdot 掳C})(70^{\circ}C) = 734500 \,\mathrm{J}\) To boil away 0.750 kg of water: \(Q_{boil} = (0.750 \,\mathrm{kg})(2.26 \times 10^6 \,\frac{J}{kg}) = 1695000 \,\mathrm{J}\) Total heat transfer required: \(Q_{total} = 47250 \,\mathrm{J} + 734500 \,\mathrm{J} + 1695000 \,\mathrm{J} = 2464750 \,\mathrm{J}\) (b) Calculating the time taken: \(t = \frac{2464750 \,\mathrm{J}}{500 \,\mathrm{W}} = 4929.5 \,\mathrm{s}\) The total heat transfer required is 2464750 J, and it will take 4929.5 seconds (approx. 82.15 minutes) for the heat transfer to occur at a rate of 500 W.

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Key Concepts

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

Understanding Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It encompasses the principles that govern the transformation of energy from one form to another and the direction in which these transformations occur. In the context of our exercise, thermodynamics helps us understand how the heat energy required to increase the temperature of the aluminum pot and the water, and to vaporize a portion of the water, is calculated and governed by thermodynamic principles.

Moreover, thermodynamics introduces us to the concept of system and surroundings. Our system in this example is the pot with water, and everything outside of this, including the heat source, is the surroundings. The heat transfer from the surroundings to the system occurs until equilibrium is reached, which, for the water, is the boiling point, at which temperature it undergoes a phase change from liquid to vapor.
Specific Heat Capacity
Specific heat capacity, often just called specific heat, is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. It is a substance-specific value, meaning that each material has its own specific heat. This concept is pivotal in calculating the amount of heat needed to increase the temperature of a given mass of material.

For instance, in our exercise, we used the specific heat of aluminum and water to calculate the heat required to increase their temperature. Aluminum's lower specific heat capacity compared to water means it takes less heat to raise the temperature of a given mass of aluminum by one degree Celsius. To put these calculations into practice, knowing the specific heats of individual substances helps us predict how they will respond to heat transfer in real-world applications.
Latent Heat of Vaporization
The latent heat of vaporization is the amount of heat energy required to change one kilogram of a substance from a liquid to a gaseous state at its boiling point, without changing temperature. It's an intensive property, meaning it doesn't depend on the mass of the substance present. However, the total heat energy required is proportional to the mass undergoing the phase change.

In relation to our problem, we calculate the heat needed to boil away a portion of the water using its latent heat of vaporization, which is quite high for water due to the strong hydrogen bonds between water molecules. As these bonds are broken, the water molecules are able to escape into the air as steam, and this transition requires a significant amount of energy.
Energy Conservation in Heat Transfer
The law of energy conservation states that energy cannot be created or destroyed, only converted from one form to another. When dealing with heat transfer, this law implies that all the heat energy provided to the system must be accounted for, whether it's used to raise the temperature or induce a phase change of the substances involved.

In the exercise, we summed the heat required for raising the temperature of both the pot and the water and the heat required for vaporizing part of the water to find the total heat transfer. This total represents the conversion of electrical energy (from the heating element) into thermal energy within the system. The concept ensures that if we measured the heat provided by the heating element, it would equal the heat absorbed by the pot and the water, adhering to the conservation of energy principle.

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Most popular questions from this chapter

Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of \(500-^{\circ} \mathrm{C}\) granite must be placed in \(4.00 \mathrm{kg}\) of \(15.0-^{\circ} \mathrm{C}\) water to bring its temperature to \(100^{\circ} \mathrm{C}\), if \(0.0250 \mathrm{kg}\) of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings.

Let's stop ignoring the greenhouse effect and incorporate it into the previous problem in a very rough way. Assume the atmosphere is a single layer, a spherical shell around Earth, with an emissivity \(e=0.77\) (chosen simply to give the right answer) at infrared wavelengths emitted by Earth and by the atmosphere. However, the atmosphere is transparent to the Sun's radiation (that is, assume the radiation is at visible wavelengths with no infrared), so the Sun's radiation reaches the surface. The greenhouse effect comes from the difference between the atmosphere's transmission of visible light and its rather strong absorption of infrared. Note that the atmosphere's radius is not significantly different from Earth's, but since the atmosphere is a layer above Earth, it emits radiation both upward and downward, so it has twice Earth's area. There are three radiative energy transfers in this problem: solar radiation absorbed by Earth's surface; infrared radiation from the surface, which is absorbed by the atmosphere according to its emissivity; and infrared radiation from the atmosphere, half of which is absorbed by Earth and half of which goes out into space. Apply the method of the previous problem to get an equation for Earth's surface and one for the atmosphere, and solve them for the two unknown temperatures, surface and atmosphere. a. In terms of Earth's radius, the constant \(\sigma,\) and the unknown temperature \(T_{s}\) of the surface, what is the power of the infrared radiation from the surface? b. What is the power of Earth's radiation absorbed by the atmosphere? c. In terms of the unknown temperature \(T_{e}\) of the atmosphere, what is the power radiated from the atmosphere? d. Write an equation that says the power of the radiation the atmosphere absorbs from Earth equals the power of the radiation it emits. e. Half of the power radiated by the atmosphere hits Earth. Write an equation that says that the power Earth absorbs from the atmosphere and the Sun equals the power that it emits. f. Solve your two equations for the unknown temperature of Earth.

The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a \(3.00-\mathrm{m}^{2}\) window that is \(0.634 \mathrm{cm}\) thick \((1 / 4 \text { in. })\) if the temperatures of the inner and outer surfaces are \(5.00^{\circ} \mathrm{C}\) and \(-10.0^{\circ} \mathrm{C},\) respectively. (This rapid rate will not be maintained - the inner surface will cool, even to the point of frost formation.)

If you place \(0^{\circ} \mathrm{C}\) ice into \(0^{\circ} \mathrm{C}\) water in an insulated container, what will the net result be? Will there be less ice and more liquid water, or more ice and less liquid water, or will the amounts stay the same?

In 1701 , the Danish astronomer Ole Romer proposed a temperature scale with two fixed points, freezing water at 7.5 degrees, and boiling water at 60.0 degrees. What is the boiling point of oxygen, \(90.2 \mathrm{K}\), on the Romer scale?
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