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Chapter 22: Problem 73

A microwave oven can heat 350 g of water from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\) in 2.00 min. (a) At what rate is energy absorbed by the water? (b) Microwaves pass through a waveguide of cross-sectional area \(88.0 \mathrm{cm}^{2} .\) What is the average intensity of the microwaves in the wave guide? (c) What are the rms electric and magnetic fields inside the wave guide?

Short Answer

Expert verified
Question: Calculate the rate of energy absorbed by the water, the average intensity of the microwaves, and the rms electric and magnetic fields inside the waveguide when a 350g cup of water is heated from 25°C to 100°C in a microwave oven with a cross-sectional area of 88.0 cm² in 2.00 minutes. Answer: Follow the steps outlined in the solution to calculate the rate of energy absorbed by the water, the average intensity of the microwaves, and the rms electric and magnetic fields inside the waveguide. Make sure to use the correct values and units for each calculation.

Step by step solution

01

Calculate the energy absorbed by the water

First, we need to determine the amount of energy absorbed by the water due to its temperature increase from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\). The energy absorbed can be calculated using the formula \(E = mcΔT\) with m=mass of water, c=specific heat of water and ΔT=change in temperature. We have: - Mass of the water (m) = 350 g - Specific heat of water (c) = \(4.186\,\text{J/(g·°C)}\) - Initial temperature (\(T_i\)) = \(25.0^{\circ} \mathrm{C}\) - Final temperature (\(T_f\)) = \(100.0^{\circ} \mathrm{C}\) Now apply the formula for energy absorbed: \(E = mc(T_f-T_i)\)
02

Calculate the rate of energy absorbed by the water

Next, we need to find the rate at which the energy is absorbed by the water. This can be determined by dividing the total energy absorbed by the time taken to heat the water. The microwave heats the water in 2.00 minutes, or 120 seconds. Rate of energy absorbed \( = \frac{E}{\text{time}}\)
03

Calculate the average intensity of the microwaves in the waveguide

Now, we'll calculate the average intensity of microwaves in the waveguide. Intensity (I) is the rate of energy flow per unit area (A): \(I = \frac{P}{A}\) where P is the microwave power absorbed by the water (from step 2) and A is the cross-sectional area of the waveguide. The cross-sectional area A is given as \(88.0\,\mathrm{cm}^{2}\), which should be converted to \(\text{m}^2\) for SI units: \(A = 88 \times 10^{-4} \,\text{m}^2\)
04

Calculate the rms electric and magnetic fields inside the waveguide

Lastly, we need to find the rms electric (\(E_{rms}\)) and magnetic (\(B_{rms}\)) fields inside the waveguide. The relationship between the intensity (I) and the fields is given by: \(I = \frac{1}{2} \frac{cE_{rms}^2}{μ_0}\) where \(c\) is the speed of light in a vacuum (\(3 \times 10^8 \,\text{m/s}\)) and \(μ_0\) is the permeability of free space (\(4π \times 10^{-7} \,\text{T·m/A}\)). From this equation, we will find \(E_{rms}\) and \(B_{rms}\): - \(E_{rms} = \sqrt{\frac{2Iμ_0}{c}}\) - \(B_{rms} = \sqrt{\frac{2I}{cμ_0}}\)

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