How Do Microwaves Affect Water Temperature in Different Sized Containers?

[andrew410]

Let us model the electromagnetic wave in a microwave oven as a plane traveling wave moving to the left, with an intensity of 26.6 kW/m^2. An oven contains two cubical containers of small mass, each full of water. One has an edge length of 6.40 cm and the other, 12.8 cm. Energy falls perpendicularly on one face of each container. The water in the smaller container absorbs 70.0% of the energy that falls on it. The water in the larger container absorbs 91.0%. (That is, the fraction 0.3 of the incoming microwave energy passes through a 6.40-cm thickness of water, and the fraction (0.3)(0.3) = 0.09 passes through a 12.8-cm thickness.) Find the temperature change of the water in each container over a time interval of 510 s. Assume that a negligible amount of energy leaves either container by heat.

How does temperature tie in with this? I can't find any formula with intensity that deals with temperature. Any help would be great! Thanks in advance! :)

 

[Data]

You're going to need a little bit of chemistry here.

\Delta E = m c \Delta T

where \Delta E is change in energy, and m, \ c, \ \Delta T are mass, specific heat capacity, and change in temperature, respectively.

Can you relate I to \Delta E?

 

[thepatientmental]

Electromagnetic waves are a type of energy that can travel through space and matter. In the case of a microwave oven, the electromagnetic waves are used to heat up food by causing the water molecules in the food to vibrate, which produces heat. The intensity of the electromagnetic wave is a measure of the energy per unit area.

In this scenario, we have two containers of water, one with an edge length of 6.40 cm and the other with an edge length of 12.8 cm. The water in each container absorbs a certain percentage of the energy from the electromagnetic wave. The smaller container absorbs 70% of the energy, while the larger container absorbs 91% of the energy. This means that the remaining energy passes through the water and is not absorbed.

To find the temperature change of the water in each container, we can use the formula for specific heat capacity: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Since we are assuming that a negligible amount of energy leaves the containers by heat, we can use the total energy absorbed by the water as the heat energy in this formula.

Let's start with the smaller container. We know that 70% of the energy is absorbed, so the remaining 30% passes through the water. This means that only 30% of the total energy is used to heat up the water. We can calculate the total energy absorbed by multiplying the intensity (26.6 kW/m^2) by the area of the container (6.40 cm x 6.40 cm). This gives us 1.08 W. Using this as the heat energy (Q) in the formula, and assuming a mass of 1 kg for the water, we can rearrange the formula to solve for ΔT. This gives us a temperature change of approximately 1.08 K over a time interval of 510 s.

For the larger container, we can follow the same steps. 91% of the energy is absorbed, so only 9% is used to heat up the water. The total energy absorbed is 4.34 W, which gives us a temperature change of approximately 4.34 K over a time interval of 510 s.

In summary, the temperature change in the smaller container is 1.08 K and in the larger container is 4.34 K over