Calculating Time to Boil Water in a Microwave Oven | Heat + Power Homework

[yaylee]

Homework Statement

A microwave oven produces energy at a rate of P = 1219 W, all in the form of microwaves. When a cup of water is placed into this oven, 64.4% of the microwaves are absorbed by the water. If a cup containing mw = 212 grams of water (227 g equals 8 ounces) starts at temperature Tw = 19.6o C: find t, the time it will take the water to reach its boiling point.

Homework Equations

Q = mcΔT
P = dW/dt

The Attempt at a Solution

Heat needed to raise 0.212 kg of water from 19.6 C to 100 C = (0.212)(4186)(100-19.6)
Q = 71349.53 Joules

Rate of Energy Absorption of Water = (64.4/100)(1219 Joules/sec) = 785 Joules/sec

Since water is absorbing heat at 785 Joules per second, in order to absorb 71349.53 Joules altogether, we need:
t = 785 J/s / 71349.53 J = 0.011 seconds.

Am I approaching the problem incorrectly?

Many thanks in advance!

 

[gneill]

Homework Statement

A microwave oven produces energy at a rate of P = 1219 W, all in the form of microwaves. When a cup of water is placed into this oven, 64.4% of the microwaves are absorbed by the water. If a cup containing mw = 212 grams of water (227 g equals 8 ounces) starts at temperature Tw = 19.6o C: find t, the time it will take the water to reach its boiling point.

Homework Equations

Q = mcΔT
P = dW/dt

The Attempt at a Solution

Heat needed to raise 0.212 kg of water from 19.6 C to 100 C = (0.212)(4186)(100-19.6)
Q = 71349.53 Joules

Rate of Energy Absorption of Water = (64.4/100)(1219 Joules/sec) = 785 Joules/sec

Since water is absorbing heat at 785 Joules per second, in order to absorb 71349.53 Joules altogether, we need:
t = 785 J/s / 71349.53 J = 0.011 seconds.

Am I approaching the problem incorrectly?

Many thanks in advance!

You're doing fine up to the last expression. Check how the units cancel.

Also, does 0.011 seconds seem a reasonable amount of time to heat up that quantity of water in a microwave?

 

[yaylee]

Thanks Gneill!

Also, just curious: do you guys help out students (like myself) voluntarily? If so , that is awfully nice !

 

[gneill]

Thanks Gneill!

Also, just curious: do you guys help out students (like myself) voluntarily? If so , that is awfully nice !

You're welcome. Yes, it's a voluntary activity and we're happy to help!

 

[Nickie]

Your approach to the problem is correct. However, there are a few things to consider in this scenario. First, you need to take into account the efficiency of the microwave oven. Not all of the 1219 W of energy produced by the microwave is actually used to heat the water. Some of it is lost due to inefficiencies in the system. Therefore, you may need to adjust your calculation by multiplying the power by the oven's efficiency.

Secondly, the rate of energy absorption may also vary depending on the temperature of the water. As the water heats up, it may absorb energy at a slower rate. This effect is known as the specific heat capacity of water. To account for this, you may need to use an average rate of energy absorption over the entire heating process.

Finally, keep in mind that the time calculated in this scenario is only an approximation. The actual time it takes for the water to reach boiling point may vary due to external factors such as the starting temperature of the water, the shape and material of the cup, and the specific settings of the microwave oven.

Overall, your approach is correct, but it is important to consider these factors to get a more accurate calculation of the time it takes for water to boil in a microwave oven.