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Chapter 13: Problem 15

A pot of water is boiling under one atmosphere of pressure. Assume that heat enters the pot only through its bottom, which is copper and rests on a heating element. In two minutes, the mass of water boiled away is \(m=0.45 \mathrm{kg} .\) The radius of the pot bottom is \(R=6.5 \mathrm{cm},\) and the thickness is \(L=2.0 \mathrm{mm} .\) What is the temperature \(T_{\mathrm{E}}\) of the heating element in contact with the pot?

Short Answer

Expert verified
The temperature of the heating element is approximately 118.5 掳C.

Step by step solution

01

Understanding the Problem

We need to determine the temperature of the heating element in contact with the pot, given that a mass of water evaporates in a certain amount of time, and the pot's bottom consists of copper with specific dimensions. We'll apply the concepts of heat transfer and the latent heat of vaporization.
02

Calculating Heat Required to Vaporize Water

The amount of heat required to vaporize the water is given by the equation \( Q = mL_v \), where \( m = 0.45 \text{ kg} \) is the mass of the water and \( L_v = 2.26 \times 10^6 \text{ J/kg} \) is the latent heat of vaporization for water. Substituting the values, we calculate \( Q = 0.45 \times 2.26 \times 10^6 = 1.017 \times 10^6 \) Joules.
03

Determine Helming Power

The heat was transferred in 120 seconds (2 minutes), so the power (rate of heat transfer) is \( P = \frac{Q}{t} = \frac{1.017 \times 10^6}{120} = 8,475 \text{ W} \).
04

Relate Power to Temperature Difference

The power can also be expressed using Fourier's law of heat conduction as \( P = \frac{kA(T_E - T_0)}{L} \), where \( k = 401 \text{ W/m路K} \) is the thermal conductivity of copper, \( A = \pi R^2 = \pi (0.065)^2 \) is the area of the pot's base, \( L = 0.002 \text{ m} \) is the thickness, and \( T_0 = 100 \text{ 掳C} \) is the boiling point of water.
05

Solving for Temperature of the Hotplate

Substitute all known values into the equation: \[ 8,475 = \frac{401 \times \pi \times (0.065)^2 \times (T_E - 100)}{0.002} \].Solve for \( T_E \). First, calculate the area: \[ A = \pi \times (0.065)^2 = 0.0133 \text{ m}^2 \].Then calculate the remaining expression: \[ 8,475 = \frac{401 \times 0.0133 \times (T_E - 100)}{0.002} \].This simplifies to: \[ 8,475 \times 0.002 = 401 \times 0.0133 \times (T_E - 100) \].Solve for \( T_E \):\[ T_E = \frac{8,475 \times 0.002}{401 \times 0.0133} + 100 \].Calculate the value to get \( T_E \approx 118.5 \text{ 掳C} \).

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Key Concepts

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

Latent Heat of Vaporization
The concept of latent heat of vaporization is crucial in understanding how much energy is required to transform a liquid into a gas. This process happens without changing the temperature of the substance. For water, the latent heat of vaporization is significantly high, at approximately 2.26 million Joules per kilogram. When you boil water, the energy supplied is used to break the intermolecular bonds, allowing the water molecules to escape as vapor.
In the original problem, we calculated the amount of heat needed to vaporize 0.45 kg of water. By multiplying the water's mass by the latent heat of vaporization, we found that approximately 1,017,000 Joules were necessary to turn the water into steam over two minutes.
This concept is not only theoretical; it has practical applications, such as in engines and cooling systems, where efficient heat management is crucial. Understanding these principles can also help in environmental studies, as the water cycle is heavily driven by heat variations and phase changes.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It describes how efficiently heat is transferred through a material when there is a temperature difference. Materials with high thermal conductivity, like metals, are excellent conductors of heat.
In our example, the copper bottom of the pot is in direct contact with the heating element and plays a key role. Copper has a thermal conductivity of 401 W/m路K, indicating that it is highly effective in transferring heat to the water above.
This property ensures that the heat from the heating element is efficiently conducted through the pot's base, allowing the water to reach its boiling point quickly. Understanding thermal conductivity is essential in the design of heating and cooling systems, ensuring that materials used can efficiently transfer heat or act as insulators when necessary.
Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction establishes a direct relationship between the heat flux through a material and the temperature gradient across it. The law is expressed as:\[ P = \frac{kA(T_E - T_0)}{L} \]Where:
  • \( P \) is the power or rate of heat transfer,
  • \( k \) is the thermal conductivity of the material,
  • \( A \) is the surface area through which heat is conducted,
  • \( T_E \) is the temperature of the heat source, and
  • \( T_0 \) is the temperature of the medium being heated,
  • \( L \) is the thickness of the material.
In the given problem, Fourier's Law was essential to find the temperature of the heating element. By knowing the thermal conductivity of copper and the dimensions of the pot's bottom, we calculated the precise amount of temperature difference needed to transfer the required heat to boil the water.
This law is foundational in thermal engineering and is crucial for designing systems requiring precise temperature control, such as electronics cooling, building insulation, and even the creation of everyday household appliances.

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Most popular questions from this chapter

Suppose the skin temperature of a naked person is \(34^{\circ} \mathrm{C}\) when the person is standing inside a room whose temperature is \(25^{\circ} \mathrm{C} .\) The skin area of the individual is \(1.5 \mathrm{m}^{2}\) (a) Assuming the emissivity is 0.80 , find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie \(=4186 \mathrm{J}\) ) that are lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

In the conduction equation \(Q=(k A \Delta T) t / L,\) the combination of terms \(k A / L\) is called the conductance. The human body has the ability to vary the conductance of the tissue beneath the skin by means of vasoconstriction and vasodilation, in which the flow of blood to the veins and capillaries is decreased and increased, respectively. The conductance can be adjusted over a range such that the tissue beneath the skin is equivalent to a thickness of \(0.080 \mathrm{mm}\) of Styrofoam or \(3.5 \mathrm{mm}\) of air. By what factor (high/low) can the body adjust the conductance?

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to \(35^{\circ} \mathrm{C}\) from \(175^{\circ} \mathrm{C}\), its net radiant power decreases to \(12.0 \mathrm{W}\). What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at \(22^{\circ} \mathrm{C}\) as the dish cools down.

The block in the drawing has dimensions \(L_{0} \times 2 L_{0} \times 3 L_{0},\) where \(L_{0}=0.30 \mathrm{m} .\) The block has a thermal conductivity of \(250 \mathrm{J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) In drawings \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is \(35^{\circ} \mathrm{C}\) and that of the cooler surface is \(19^{\circ} \mathrm{C} .\) Determine the heat that flows in \(5.0 \mathrm{s}\) for each case.

Two pots are identical except that the flat bottom of one is aluminum, whereas that of the other is copper. Water in these pots is boiling away at \(100.0^{\circ} \mathrm{C}\) at the same rate. The temperature of the heating element on which the aluminum bottom is sitting is \(155.0^{\circ} \mathrm{C} .\) Assume that heat enters the water only through the bottoms of the pots and find the temperature of the heating element on which the copper bottom rests.
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