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

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The discussion focuses on calculating the time required to boil water in a microwave oven, given a power output of 1219 W and a water absorption rate of 64.4%. The heat needed to raise 212 grams of water from 19.6°C to 100°C is calculated to be approximately 71349.53 Joules. The rate of energy absorption by the water is determined to be 785 Joules per second. However, a calculation error is pointed out, as the resulting time of 0.011 seconds is deemed unrealistic for heating that amount of water. The participants confirm their willingness to assist students voluntarily.
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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!
 
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yaylee said:

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? :rolleyes:
 
Thanks Gneill!

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