Problem 53 - College Physics

Question

A pot with a steel bottom \(8.50 \mathrm{~mm}\) thick rests on a hot stove. The area of the bottom of the pot is \(0.150 \mathrm{~m}^{2}\). The water inside the pot is at \(100.0^{\circ} \mathrm{C}\), and \(0.390 \mathrm{~kg}\) are evaporated every \(3.00 \mathrm{~min}\). Find the temperature of the lower surface of the pot, which is in contact with the stove.

Step-by-Step Solution

Answer

The lower surface temperature is approximately 112°C.

Step 1: Understand the Problem

We need to find the temperature at the lower surface of the pot. Given data: steel bottom thickness, area, rate of water evaporation, water boiling point, and time period of evaporation.

Step 2: Calculate the Heat Required for Evaporation

The latent heat of vaporization for water is typically \( L = 2260 \text{ kJ/kg} \). Since \(0.390 \text{ kg}\) of water is evaporated every \(3.00 \text{ min}\), the heat required per minute, \( Q \), in \( ext{J} \) is calculated as: \[ Q = m \times L = 0.390 \times 2260000 \].

Step 3: Use Heat Transfer Equation

Heat conduction through the pot's bottom follows Fourier’s Law: \( Q = k \cdot A \cdot \frac{T_1 - T_2}{d} \), where \( A = 0.150 \, \text{m}^2 \) (area), \( d = 8.5 \, \text{mm} = 0.0085 \, \text{m} \) (thickness), and \( k \) is thermal conductivity of steel \( = 50 \, \text{W/m} \cdot \text{K} \).

Step 4: Calculate Heat Transfer Rate per Second

Convert the calculated heat from Step 2 to per second (since \(3.00 \, ext{min} = 180 \, ext{s} \)): \[ \frac{Q}{3 imes 60} = \frac{0.390 \times 2260000}{180} \text{ W} \].

Step 5: Solve for the Temperature Difference

Substitute all values into Fourier’s Law and solve for \( T_1 - T_2 \): \[ Q' = 50 \cdot 0.150 \cdot \frac{T_1 - 100}{0.0085} \]. Rearrange and solve for \( T_1 \).

Step 6: Calculate Temperature of Lower Surface

From the previous steps, \( T_1 \), the temperature of the lower surface of the pot, is calculated as follows: solve for \( T_1 \); \[ T_1 = Q' \times \frac{0.0085}{50 \times 0.150} + 100 \].

Key Concepts

Latent Heat of Vaporization

The latent heat of vaporization is an essential concept in thermal physics, particularly when studying phase changes. It refers to the amount of heat required to convert a unit mass of a liquid to vapor without a change in temperature. For water, the latent heat of vaporization is generally around \(2260 \, \text{kJ/kg}\), meaning that this amount of heat is necessary to evaporate just one kilogram of water.
Understanding this concept is crucial when examining processes like boiling, where energy input must continuously compensate for the heat absorbed by the liquid. In our exercise, the evaporation of \(0.390\, \text{kg}\) of water involved calculating the total heat required using the latent heat formula \(Q = m \times L\), thereby providing a basis for further calculations on heat transfer across the pot's material.

Fourier's Law

Fourier's Law explains the conduction of heat across materials. It states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the area through which the heat flows. The mathematical representation of Fourier's Law is \( Q = k \times A \times \frac{T_1 - T_2}{d} \), where:

• \(k\) is the thermal conductivity of the material.
• \(A\) is the area through which heat is conducted.
• \(T_1\) and \(T_2\) are the temperatures across the material's thickness.
• \(d\) is the thickness of the material.

For the pot on a stove, Fourier's Law is used to calculate the heat transfer rate needed to maintain the water at its boiling point. Given that the steel bottom has a known thickness and area, the law helps to find the temperature of the pot's lower surface in contact with the stove.

Thermal Conductivity

Thermal conductivity is a material-specific property indicating how well heat is conducted through the material. It is denoted by \(k\) in Fourier's Law and is measured in \(\text{W/m} \cdot \text{K}\).
In the exercise, the thermal conductivity of steel is taken as \(50 \, \text{W/m} \cdot \text{K}\). This value is crucial because it determines how effective the pot's bottom is in transferring heat from the stove to the water. A higher thermal conductivity means more efficient heat transfer, resulting in faster heating of the water. The choice of material greatly impacts the energy efficiency and speed of cooking, making thermal conductivity an important factor in cookware design.

Boiling Point of Water

The boiling point of water is a fundamental reference point in thermal studies and cooking applications. At sea level, water boils at \(100.0^{\circ} \text{C}\).
In the given exercise, maintaining water at its boiling point is necessary for evaporating the specific quantity within the provided time. This means any heat applied has to ensure that sufficient energy is available to sustain the boiling process, enabling the water to transition from liquid to vapor.
Understanding the boiling point is crucial, as it informs the energy requirements in heating and helps in controlling cooking conditions, ensuring effective heat management. The boiling point will slightly vary at different altitudes, since atmospheric pressure influences when water molecules gain enough energy to phase change.