Problem 71 - University Physics with Modern 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 temperature of the lower surface of the pot is approximately \(106^{\circ}C\).

Step 1: Understand the Energy Transfer Mechanism

The pot transfers heat through conduction from the stove to the water. The heat energy is used to evaporate the water. We can use the formula for conductive heat transfer, which is governed by the equation: \[ Q = \frac{k \cdot A \cdot (T_{ ext{stove}} - T_{ ext{water}})}{d} \cdot t \]where \(Q\) is the heat transfer, \(k\) is the thermal conductivity of the steel bottom, \(A\) is the area, \(d\) is the thickness of the steel, and \(t\) is the time.

Step 2: Calculate the Heat Required to Evaporate Water

The heat required to evaporate water can be found using the latent heat of vaporization. The formula is:\[ Q = m \cdot L \]where \(m\) is the mass of water evaporated (0.390 kg) and \(L\) is the latent heat of vaporization for water (2260 kJ/kg).Converting \(L\) to J/kg for consistency:\[ L = 2260 \times 10^3 \, \text{J/kg} \]Thus,\[ Q = 0.390 \, \text{kg} \times 2260 \times 10^3 \, \text{J/kg} = 8.814 \times 10^5 \, \text{J} \]

Step 3: Relate Heat Transfer via Conduction

The same heat \(Q\) is conducted through the pot's base from the stove to the water. Given:- Conductive heat transfer equation:\[ Q = \frac{k \cdot A \cdot (T_{ ext{stove}} - T_{ ext{water}})}{d} \cdot t \]- Known variables: - \(Q = 8.814 \times 10^5 \, \text{J}\) - \(A = 0.150 \, \text{m}^2\) - \(d = 8.50 \, \text{mm} = 0.0085 \, \text{m}\) - \(t = 3 \, \text{min} = 180 \, \text{s}\) - \(T_{ ext{water}} = 100^{\circ}C\) - \(k = 46 \, \text{W/m} \cdot \text{K}\) for typical steel

Step 4: Solve for Stove Temperature

By substituting back into the heat conduction formula:\[ 8.814 \times 10^5 = \frac{46 \times 0.150 \times (T_{ ext{stove}} - 100)}{0.0085} \times 180 \]Now, solve for \(T_{\text{stove}}\):1. Simplify the right side:\[ 8.814 \times 10^5 = \frac{46 \times 27 \times (T_{ ext{stove}} - 100)}{0.0085} \]2. Rearrange and solve:\[ 8.814 \times 10^5 = 145800(T_{ ext{stove}} - 100) \]\[ T_{\text{stove}} - 100 = \frac{8.814 \times 10^5}{145800} \]\[ T_{\text{stove}} = 100 + 6.04 \]\[ T_{\text{stove}} \approx 106.04^{\circ}C \]

Step 5: Conclusion

The temperature of the lower surface of the pot in contact with the stove is approximately \(106^{\circ}C\).

Key Concepts

Conduction

Conduction is a fundamental mechanism through which heat is transferred. When two surfaces are in contact, heat moves from the hotter object to the cooler one. In the case of the pot on the stove, heat is conducted through the steel bottom from the stove to the water within the pot.

This process occurs because of the temperature difference between the stove and the water. The stove is hotter, and as dictated by the laws of physics, heat naturally transfers towards the cooler region, i.e., the water. The rate at which the heat is transferred depends on several factors: the thermal conductivity of the material, the temperature gradient, and the thickness of the material. The thinner the material, the faster the heat transfer, which is described by Fourier's Law of heat conduction. Here, we utilize this concept to calculate the amount of heat required to evaporate a given mass of water in the pot.

Thermal Conductivity

Thermal conductivity is a property of a material that indicates its ability to conduct heat. In simpler terms, it measures how well heat passes through a substance. For example, metals generally have high thermal conductivity, which means they are good at transferring heat quickly.

In this exercise, the pot has a steel bottom with a thermal conductivity of 46 W/m·K. This is relatively high, signifying that steel is efficient at conducting heat, making it suitable for applications like cooking pots. The value of thermal conductivity plays a crucial role in determining the rate of heat transfer through the pot, alongside the thickness and surface area of the pot's bottom.

• A higher thermal conductivity value means more heat can be transferred through the pot in a given time.
• The area and thickness of the pot's bottom also impact how heat is conducted.

Understanding these relationships allows us to calculate how much heat is transferred from the stove to the water, helping solve physics problems involving heating and cooling.

Latent Heat of Vaporization

Latent heat of vaporization is the amount of heat required to turn a liquid into a gas without changing its temperature. This concept is crucial when dealing with processes like boiling or evaporation. In this problem, water at 100°C is transformed into steam, which requires a specific amount of energy called the latent heat of vaporization.

For water, this heat is quite high, at 2260 kJ/kg, reflecting the energy needed to break the molecular bonds holding the liquid together. To determine the energy required to evaporate the given mass of water, we use the formula \( Q = m \cdot L \), where \( m \) is the mass of water, and \( L \) is the latent heat of vaporization.

This energy computation is essential for understanding how much heat the stove must supply to convert the water into steam, aiding in accurately solving thermal conduction exercises in physics.

Problem Solving in Physics

Problem solving in physics often involves breaking down a complex phenomenon into understandable segments. This particular exercise showcases how to dissect a practical scenario involving heat transfer through the concepts of conduction, thermal conductivity, and latent heat.

The first step is understanding the mechanisms involved, such as how heat is transferred through conduction. Identifying known variables and unknowns helps set up the appropriate equations based on these underlying physics principles.

By organizing information clearly and applying suitable formulas, such as those for conduction and latent heat, we can then calculate variables step-by-step.

• Identify all given quantities and any constants, like thermal conductivity and latent heat values.
• Apply relevant equations to relate these quantities.
• Carefully rearrange equations to solve for the unknowns.

Tying these steps together is crucial, allowing for a comprehensive solution to real-world physics problems.