Problem 69 - 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 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 is found by adding the calculated \( \Delta T \) to \( 100.0^{\circ}C \).

Step 1: Understanding the Heat Transfer

The problem involves heat conduction through the steel bottom from the stove to the pot. The equation we use is the thermal conduction formula: \[ q = \frac{k \cdot A \cdot \Delta T}{d} \]where \( q \) is the rate of heat transfer (W), \( k \) is the thermal conductivity of steel, \( A \) is the area, \( \Delta T \) is the temperature difference, and \( d \) is the thickness of the steel.

Step 2: Calculating Heat Required for Evaporation

The amount of heat needed to evaporate a certain mass of water can be found using the equation: \[ q = m \cdot L_v \]where \( m \) is the mass of the water evaporated (kg) and \( L_v \) is the latent heat of vaporization of water. For water, \( L_v \) is approximately \( 2260 \times 10^3 \) J/kg. Given \( m = 0.390 \) kg, calculate \( q \).

Step 3: Calculating Rate of Heat Transfer

Convert the heat calculated for evaporation into a power (rate of heat transfer) since evaporation occurs over time. \[ \dot{q} = \frac{q}{time} \]Here, time is 3 minutes, converted into seconds: \( 3 \times 60 = 180 \) seconds. Substitute the value of \( q \) and divide by 180 to find \( \dot{q} \) in watts.

Step 4: Applying the Thermal Conduction Formula

Rearrange the thermal conduction formula to solve for \( \Delta T \):\[ \Delta T = \frac{\dot{q} \cdot d}{k \cdot A} \]Use the known values: thermal conductivity of steel \( k = 50 \) W/m°C, thickness \( d = 8.50 \times 10^{-3} \) m, and area \( A = 0.150 \) m². Substitute these into the equation along with \( \dot{q} \) from the previous step.

Step 5: Finding the Lower Surface Temperature

Finally, calculate the temperature of the lower surface of the steel bottom \( T_{lower} \) knowing that \( T_{upper} = 100.0^{\circ}C \) (temperature of boiling water):\[ T_{lower} = T_{upper} + \Delta T \]Add the temperature difference \( \Delta T \) calculated in the previous step to \( 100.0 \)°C to find \( T_{lower} \).

Key Concepts

Thermal Conduction

Thermal conduction is the process of heat energy being transferred through a material without any movement of the material itself. Imagine heat as water flowing through a pipe, where the material is the pipe. In this exercise, thermal conduction occurs in the steel bottom of a pot on a stove. Heat moves from the stove through the steel to the water inside the pot.

The rate of heat transfer can be calculated using the formula: \[ q = \frac{k \cdot A \cdot \Delta T}{d} \] where:

• \( q \) is the rate of heat transfer in watts (W)
• 
  \( k \) is the thermal conductivity of the material. For steel, it is around 50 W/m°C
• 
  \( A \) is the area through which heat is being transferred,
• 
  \( \Delta T \) is the temperature difference across the material,
• 
  and \( d \) is the thickness of the material.

Understanding this concept is crucial for calculating the temperature of the surface in contact with the stove.

Latent Heat of Vaporization

Latent heat of vaporization refers to the heat required to convert a unit mass of a liquid into vapor without a temperature change. It is an essential concept when dealing with phase changes, such as boiling water turning into steam.

In our exercise scenario, we have water boiling and evaporating inside the pot. The amount of heat needed for this phase change can be pinpointed using the formula: \[ q = m \cdot L_v \] where:

• \( q \) is the heat required (in joules),
• 
  \( m \) is the mass of water that evaporates (in kg),
• 
  \( L_v \) is the latent heat of vaporization. For water, it is usually taken as \( 2260 \times 10^3 \) J/kg.

This concept tells us how much energy from the stove is consumed in changing water into steam, rather than just warming it up.

Temperature Difference Calculation

Temperature difference calculation is a straightforward yet critical step in determining the overall heat flow through a material. It helps to know how much hotter one side of a material is compared to the other.

To find this temperature difference, use the rearranged version of the thermal conduction formula: \[ \Delta T = \frac{\dot{q} \cdot d}{k \cdot A} \] Here, \( \dot{q} \) is the rate of heat transfer measured previously.

This formula allows us to solve for \( \Delta T \) when we have values for the rate of heat transfer, the thickness of the material, and its area. With the temperature difference, we can now calculate the temperature at the stove-pot interface, giving us a clearer picture of heat distribution.

Thermal Conductivity of Steel

Thermal conductivity is a property that defines how well a material can conduct heat. In this case, we are looking at steel, which is generally known as a good heat conductor.

For steel, the thermal conductivity is approximately 50 W/m°C. This value is crucial when using the thermal conduction formula. It tells us how easily heat can travel through the steel bottom of the pot.

You need to consider:

• The material's thermal conductivity affects the efficiency of heat transfer. Steel's high thermal conductivity means that heat from the stove quickly reaches the water in the pot.
• Understanding the thermal conductivity helps in designing better cookware that heats efficiently and evenly.

This knowledge helps in estimating how much energy is used not just to heat the water but also to allow enough room for the steel to transmit heat effectively.