/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Refer to Interactive Solution \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Refer to Interactive Solution \(\underline{13.13}\) at for help in solving this problem. In an aluminum pot, \(0.15 \mathrm{~kg}\) of water at \(100^{\circ} \mathrm{C}\) boils away in four minutes. The bottom of the pot is \(3.1 \times 10^{-3} \mathrm{~m}\) thick and has a surface area of \(0.015 \mathrm{~m}^{2}\). To prevent the water from boiling too rapidly, a stainless steel plate has been placed between the pot and the heating element. The plate is \(1.4 \times 10^{-3} \mathrm{~m}\) thick, and its area matches that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot, find the temperature at (a) the aluminum-steel interface and (b) the steel surface in contact with the heating element.

Short Answer

Expert verified
(a) The temperature at the aluminum-steel interface is \(90.41^{\circ}C\). (b) The temperature at the steel-heating element interface is \(78.25^{\circ}C\).

Step by step solution

01

Understand the problem

We need to find the temperatures at two interfaces: (a) the aluminum-steel interface and (b) the steel surface in contact with the heating element. This involves understanding heat conduction through both materials.
02

Gather necessary information

The given data includes: 1. Mass of water boiled away: 0.15 kg2. Time: 4 minutes3. Thickness of aluminum pot bottom: \(3.1 \times 10^{-3} \text{ m}\)4. Thickness of steel plate: \(1.4 \times 10^{-3} \text{ m}\)5. Surface area: \(0.015 \text{ m}^2\)6. Latent heat of vaporization of water: \(2260 \text{ kJ/kg}\)7. Thermal conductivity of aluminum: \(237 \text{ W/m} \cdot \text{K}\)8. Thermal conductivity of steel: \(16 \text{ W/m} \cdot \text{K}\)
03

Calculate the power needed to boil away water

The power needed can be found using the latent heat formula. \[ Q = mL = 0.15 \times 2260 \times 10^3 = 339,000 \text{ J} \]Given that 339,000 J of energy must be supplied in 4 minutes (or 240 seconds):\[ P = \frac{Q}{t} = \frac{339,000}{240} = 1412.5 \text{ W} \]
04

Set up heat conduction equations

For heat conduction, we have\[ P = \frac{kA(T_{hot} - T_{cold})}{d} \]Where:- \(P\) is the power.- \(k\) is the thermal conductivity.- \(A\) is the area.- \(T_{hot} - T_{cold}\) is the temperature difference.- \(d\) is the thickness.
05

Calculate temperature at the aluminum-steel interface

Using the equation for aluminum:\[ 1412.5 = \frac{237 \times 0.015 \times (100 - T_{interface})}{3.1 \times 10^{-3}} \]Solving for \(T_{interface}\):\[ T_{interface} = 100 - \frac{1412.5 \times 3.1 \times 10^{-3}}{237 \times 0.015} \]\[ T_{interface} \approx 90.41 ^{\circ} \text{C} \]
06

Calculate the temperature at the steel-heating element interface

Using the equation for steel:\[ 1412.5 = \frac{16 \times 0.015 \times (T_{interface} - T_{element})}{1.4 \times 10^{-3}} \]Plug in the \( T_{interface} \approx 90.41 ^{\circ} \text{C} \) and solve for \(T_{element}\):\[ T_{element} = 90.41 - \frac{1412.5 \times 1.4 \times 10^{-3}}{16 \times 0.015} \]\[ T_{element} \approx 78.25 ^{\circ} \text{C} \]
07

Conclude with results

The temperature at the aluminum-steel interface is approximately \(90.41^{\circ} \text{C}\), and the temperature at the steel surface in contact with the heating element is approximately \(78.25^{\circ} \text{C}\).

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Key Concepts

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

Thermal conductivity
Thermal conductivity is a crucial concept in understanding how heat moves through materials. It measures a material's ability to conduct – or transport – heat. In this exercise, both aluminum and steel have different thermal conductivities, influencing how they conduct heat from the pot to the heating element.

- **Aluminum** has a high thermal conductivity of 237 W/m·K, meaning it efficiently transfers heat. That is why it is commonly used in cookware.

- **Steel**, on the other hand, has a lower thermal conductivity of 16 W/m·K, which slows down the heat transfer.

These differences highlight the roles these materials play: aluminum transfers heat quickly to boil the water, while steel tempers the heat transfer to prevent boiling too fast.
Latent heat of vaporization
Latent heat of vaporization is the heat required to transform a given amount of liquid into vapor without changing its temperature. For water, this value is notably high, at 2260 kJ/kg, indicating it needs substantial energy to convert from liquid to gas.

In the exercise, this concept helps calculate the energy needed to boil away 0.15 kg of water. Given the latent heat:
  • We use the formula: \[ Q = mL \] where \Q\ is the heat required, \m\ is the mass, and \L\ is the latent heat of vaporization.
    This results in an energy requirement of 339,000 J to boil the water in the pot.
Understanding latent heat is essential in thermal analysis because it quantifies energy changes during phase transitions like boiling.
Temperature difference
Temperature difference is the driving force behind heat conduction. It represents the variance in temperature between two points, causing heat flow from the hotter to the cooler area.

In our context:
  • The water is initially at 100°C – a significant temperature to initiate boiling.
  • The aluminum-steel interface and steel-heating element interface have lower temperatures due to gradual heat loss as heat moves through each layer.
This steady decline in temperature illustrates how heat dissipates through materials, helping students predict and control heat transfer in similar practical applications.
Heat transfer equations
Heat transfer equations are the backbone of calculating heat flow in various situations. In this exercise, we employ the conduction equation:

- **Formula**: \[ P = \frac{kA(T_{hot} - T_{cold})}{d} \]
This formula helps find the temperatures at critical junctions between materials.
  • **Variables include**:
  • \P\: the power or heat transfer rate.
  • \k\: thermal conductivity of the material.
  • \A\: cross-sectional area.
  • \T_{hot} - T_{cold}\: temperature difference.
  • \d\: thickness of the material.
By using this formula, students can calculate precise temperature differences across materials, essential for efficient thermal management.

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Most popular questions from this chapter

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?

Concept Questions The block shown in the drawing has dimensions \(L_{0} \times 2 L_{0} \times 3 L_{0}\). In drawings \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), heat is conducted through the block in three different directions. In each case, the same temperature difference exists between the opposite surfaces through which the heat passes, and the time during which the heat flows is the same. (a) The cross-sectional area of the opposite surfaces in \(\mathrm{C}\) is greater than that in A. Does this fact alone mean that the heat conducted in \(\mathrm{C}\) is greater than that conducted in A? Provide a reason for your answer. (b) The length of material through which heat is conducted is greater in A than in B. Does this fact alone imply that the heat conducted in A is smaller than that conducted in B? Why? (c) Rank the heat conducted in each of the three cases, largest first.

In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissivity \(=0.75\) ). it has a temperature of \(62^{\circ} \mathrm{C}\). The new owner of the house paints the radiator a lighter color (emissivity \(=0.50\) ). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?

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.

The temperature in an electric oven is \(160{ }^{\circ} \mathrm{C}\). The temperature at the outer surface in the kitchen is \(50^{\circ} \mathrm{C}\). The oven (surface area \(=1.6 \mathrm{~m}^{2}\) ) is insulated with material that has a thickness of \(0.020 \mathrm{~m}\) and a thermal conductivity of \(0.045 \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) (a) How much energy is used to operate the oven for six hours? (b) At a price of \(\$ 0.10\) per kilowatt. hour for electrical energy, what is the cost of operating the oven?

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