On its highest power setting, a microwave oven increases the temperature of 0.400 kg of spaghetti by 45.0 °C in 120 s. The specific heat of spaghetti is 3760 J/kg °C and the microwaves are absorbed over a circular area 20.0 cm in diameter. What is the peak electric field strength of the microwave radiation?

1 answer

To find the peak electric field strength of the microwave radiation, we need to first find the energy absorbed by the spaghetti, and then use the energy to find the average power. Finally, we can use the power to find the peak electric field strength.

1. Calculate the energy absorbed by the spaghetti:
Energy = mass x specific heat x temperature change
Energy = 0.400 kg x 3760 J/kg °C x 45.0 °C = 67680 J

2. Calculate the average power:
Power = Energy / time
Power = 67680 J / 120 s = 564 W

3. Calculate the intensity of the microwave radiation:
Intensity is defined as the power per unit area. The area can be found using the diameter of the circular area:
Area = π x (diameter / 2)^2
Area = π x (0.20 m / 2)^2 = 0.0314 m²

Intensity = Power / Area
Intensity = 564 W / 0.0314 m² ≈ 18000 W/m²

4. Calculate the peak electric field strength:
The intensity of an electromagnetic wave is related to the peak electric field strength (E) and the speed of light (c) by the following formula:

Intensity = (ε₀ * c * E^2) / 2

Where ε₀ is the vacuum permittivity (8.854 x 10^(-12) C²/N m²).

Rearrange the formula to solve for E:

E^2 = (2 * Intensity) / (ε₀ * c)
E = sqrt((2 * Intensity) / (ε₀ * c))

Now substitute the values:

E = sqrt((2 * 18000 W/m²) / (8.854 x 10^(-12) C²/N m² * 3 x 10^8 m/s))
E ≈ 979 V/m

So the peak electric field strength of the microwave radiation is approximately 979 V/m.