Problem 55 - Physics

Question

Review Interactive Solution \(\underline{16.55}\) at for one approach to this problem. A dish of lasagna is being heated in a microwave oven. The effective area of the lasagna that is exposed to the microwaves is \(2.2 \times 10^{-2} \mathrm{~m}^{2}\). The mass of the lasagna is \(0.35 \mathrm{~kg},\) and its specific heat capacity is \(3200 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). The temperature rises by \(72 \mathrm{C}^{\circ}\) in 8.0 minutes. What is the intensity of the microwaves in the oven?

Step-by-Step Solution

Answer

The intensity of the microwaves is approximately \(7636 \text{ W/m}^2\).

Step 1: Understand the Given Data

The problem provides the following data: the effective area of lasagna exposed to microwaves is \(2.2 \times 10^{-2} \text{ m}^2\), the mass of the lasagna is \(0.35 \text{ kg}\), the specific heat capacity is \(3200 \text{ J/(kg} \cdot \text{°C)}\), the temperature rise is \(72 \text{ °C}\), and the time taken is 8.0 minutes.

Step 2: Find the Energy Required

To find the energy required to heat the lasagna, use the formula: \[ Q = mc\Delta T \] where \( m = 0.35 \text{ kg}\), \( c = 3200 \text{ J/(kg} \cdot \text{°C)}\), and \( \Delta T = 72 \text{ °C}\). Substituting the values, \[ Q = 0.35 \times 3200 \times 72 \] Calculate this to find the energy \( Q \).

Step 3: Perform the Calculation

Calculate the energy:\[ Q = 0.35 \times 3200 \times 72 = 80640 \text{ J} \]

Step 4: Convert Time to Seconds

Convert the time from minutes to seconds to get it in a standard unit. Since there are 60 seconds in a minute:\[ 8.0 \text{ minutes} = 8.0 \times 60 = 480 \text{ seconds} \]

Step 5: Calculate Power Absorbed by Lasagna

Power is the rate at which energy is transferred and is given by \[ P = \frac{Q}{t} \] where \( Q = 80640 \text{ J} \) and \( t = 480 \text{ s} \). Use these to find \( P \).

Step 6: Perform Power Calculation

Calculate the power:\[ P = \frac{80640}{480} \approx 168 \text{ W} \]

Step 7: Calculate Microwave Intensity

Intensity \( I \) is the power per unit area, given by \[ I = \frac{P}{A} \] where \( P = 168 \text{ W} \) and \( A = 2.2 \times 10^{-2} \text{ m}^2 \). Use these values to find the intensity.

Step 8: Perform Intensity Calculation

Compute the intensity:\[ I = \frac{168}{2.2 \times 10^{-2}} \approx 7636 \text{ W/m}^2 \]

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property of a material that shows how much energy is required to raise the temperature of one kilogram of the substance by one degree Celsius. In this exercise, the lasagna has a specific heat capacity of 3200 J/(kg°C). This means it needs 3200 joules of energy to increase the temperature of one kilogram of lasagna by 1°C.
Understanding specific heat capacity helps us calculate how much energy is needed to heat a certain mass of lasagna by a specified temperature change. We use the formula:

• \( Q = mc\Delta T \)

where:

• \( Q \) is the energy required
• \( m \) is the mass of the lasagna
• \( c \) is the specific heat capacity
• \( \Delta T \) is the change in temperature

This formula allows us to calculate that the energy required to heat the lasagna by 72°C is 80640 joules by substituting the given values.

Microwave Heating

Microwave heating involves the use of microwaves, a type of electromagnetic wave, to heat objects. These waves cause water molecules in the food to vibrate, producing heat. In this scenario, the lasagna is heated by microwaves that penetrate it through an effective area. The effective area here is provided as 2.2 \( \times 10^{-2} \) m².
The heating mechanism is efficient and allows the food to heat up quickly, as the energy is directly transferred to the water molecules within the lasagna. Understanding how microwaves work is crucial to determining how efficiently the energy is absorbed by the food, which ultimately influences the intensity calculation.

Energy Transfer

Energy transfer in the context of heating involves the movement of energy from the microwave oven to the lasagna. The energy is absorbed by the lasagna's molecules, increasing their kinetic energy and thus the temperature of the lasagna.
To find how much energy was transferred in this exercise, we compute the total energy necessary to heat the lasagna using the formula for specific heat capacity. In our calculations, we found that the energy required (\( Q \)) is 80640 joules. This energy is transferred over a period of time (8 minutes, converted to seconds, which is 480 seconds) to the lasagna, giving a power output using the formula:

• \( P = \frac{Q}{t} \)

where:

• \( Q \) is the energy transferred
• \( t \) is the time in seconds

Calculations show the power absorbed by the lasagna is approximately 168 watts.

Intensity Calculation

Intensity is an important concept in understanding how microwaves function, especially when considering how much power the lasagna absorbs per unit area. Intensity (\( I \)) is calculated as the power absorbed divided by the effective area (\( A \)) exposed to the microwaves.The formula is:

• \( I = \frac{P}{A} \)

where:

• \( P \) is the power absorbed (168 watts in our scenario)
• \( A \) is the area through which the microwaves penetrate the lasagna (2.2 \( \times 10^{-2} \) m²)

By substituting these values into the intensity formula, we find that the intensity of the microwaves in the oven is approximately 7636 watts per square meter. This high intensity ensures that the lasagna heats up relatively quickly, as more energy is concentrated over a small area, leading to efficient heating.