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Chapter 13: Problem 10

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.

Short Answer

Expert verified
The temperature of the heating element for the copper bottom is approximately \(132.5^{\circ}C\).

Step by step solution

01

Understand Heat Transfer Equation

The rate of heat transfer, known as the heat flux, through a material is given by the formula \( Q = \dfrac{k \cdot A \cdot (T_{hot} - T_{cold})}{d} \), where \( Q \) is the heat transfer rate, \( k \) is the thermal conductivity of the material, \( A \) is the area, \( T_{hot} \) and \( T_{cold} \) are the temperatures of the hot and cold sides respectively, and \( d \) is the thickness of the bottom.
02

Equate Heat Transfer Rates for Both Pots

Since the water in both pots is boiling at the same rate, the heat transfer rates for both pots are equal. Express this equality as \( \dfrac{k_{Al} \cdot A \cdot (T_{heater,Al} - 100)}{d_{Al}} = \dfrac{k_{Cu} \cdot A \cdot (T_{heater,Cu} - 100)}{d_{Cu}} \).
03

Eliminate Cross-sectional Area and Thickness

The bottoms of the pots are identical, so the area \( A \) and thickness \( d \) cancel out. Therefore, the equation reduces to \( k_{Al} \cdot (T_{heater,Al} - 100) = k_{Cu} \cdot (T_{heater,Cu} - 100) \).
04

Insert Known Values

Insert the known values: \( T_{heater,Al} = 155.0^{\circ}C \), \( k_{Al} = 237 \text{W/mK} \) (thermal conductivity of aluminum), and \( k_{Cu} = 401 \text{W/mK} \) (thermal conductivity of copper). The equation becomes \( 237 \cdot (155 - 100) = 401 \cdot (T_{heater,Cu} - 100) \).
05

Solve for the Unknown Temperature

Simplify and solve for \( T_{heater,Cu} \). Start with calculating \( 237 \cdot 55 = 13035 \). Then, divide by the copper's thermal conductivity: \( T_{heater,Cu} - 100 = \dfrac{13035}{401} \), resulting in \( T_{heater,Cu} - 100 = 32.52 \). Thus, \( T_{heater,Cu} \approx 132.52^{\circ}C \).

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Key Concepts

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

Understanding Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It tells us how efficiently heat can pass through a substance.
High thermal conductivity means heat can move quickly through the material, whereas low thermal conductivity means the material holds onto heat for longer.
In this exercise, we are comparing thermal conductivities of aluminum and copper, which are two common materials used in cookware.
  • Copper has a higher thermal conductivity (401 W/mK) than aluminum (237 W/mK).
  • This means copper heats up and cools down faster, making it more efficient for transferring heat.
  • The efficiency of heat transfer is crucial in boiling processes, as it influences how quickly water reaches its boiling point.
Understanding the thermal conductivity of materials helps in designing and choosing the right cookware for efficient cooking.
The Concept of Boiling Point
The boiling point of a substance is the temperature at which it changes from liquid to gas. For water, this temperature is typically 100掳C at sea level.
In this exercise, the boiling point of water is a critical element because both pots have water that is boiling away at the same rate.
To maintain boiling, the heat supplied must be consistent and sufficient to sustain the phase change from liquid to vapor.
  • The boiling point can change with atmospheric pressure, but here it is a constant since the exercise specifies boiling water.
  • This steady-state condition means the heat energy going into both pots must be the same to keep the water boiling consistently.
  • We assume that heat enters the water only through the bottoms of the pots.
Knowing about the boiling point helps us understand the importance of the constant heat input necessary to keep a substance in its boiling state.
What is Temperature Gradient?
Temperature gradient is the rate at which temperature changes in a specific direction.In the context of this exercise, it refers to the difference in temperature between the heating element and the boiling water's surface. This difference drives the flow of heat from the heating element into the water.
  • The larger the temperature gradient, the more intense the heat transfer between two surfaces, which determines how quickly the heat reaches the boiling water.
  • For the pots in the exercise, the gradient is measured between the heating element's surface and the boiling point of the water (100掳C).
  • In the equation, it appears as \((T_{heater} - 100)\)", showing the temperature gradient for both aluminum and copper pots.
By understanding the temperature gradient, you can comprehend how effectively a pot transfers heat and how that affects cooking efficiency.

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Most popular questions from this chapter

The filament of a light bulb has a temperature of \(3.0 \times 10^{3}{ }^{\circ} \mathrm{C}\) and radiates sixty watts of power. The emissivity of the filament is \(0.36 .\) Find the surface area of the filament.

Two cylindrical rods have the same mass. One is made of silver (density \(=10\) \(\left.500 \mathrm{~kg} / \mathrm{m}^{3}\right)\), and one is made of iron (density \(\left.=7860 \mathrm{~kg} / \mathrm{m}^{3}\right)\). Both rods conduct the same amount of heat per second when the same temperature difference is maintained across their ends. What is the ratio (silver-to-iron) of (a) the lengths and (b) the radii of these rods?

A pot of water is boiling on a stove 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 a certain time, a mass \(m\) of water boils away. (a) What is the temperature of the boiling water and does it change during this time? (b) What determines the amount of heat needed to boil the water? (c) Is the temperature of the heating element in contact with the pot greater than, smaller than, or equal to \(100{ }^{\circ} \mathrm{C} ?\) Explain. Problem 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? Verify that your answer is consistent with your answers to the Concept Questions.

A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders, each of which has the same length. Each small cylinder has the same temperature as the original cylinder. The total radiant power emitted by the pieces is twice that emitted by the original cylinder. How many smaller cylinders are there?

How many days does it take for a perfect blackbody cube \((0.0100 \mathrm{~m}\) on a side, \(30.0^{\circ} \mathrm{C}\) ) to radiate the same amount of energy that a one-hundred-watt light bulb uses in one hour?
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