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

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Homework Help Overview

The discussion revolves around calculating the time required to boil water in a microwave oven, focusing on the energy absorption rate and the specific heat capacity of water. The problem involves understanding the relationship between power, energy, and temperature change.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of heat energy required to raise the temperature of water and the rate at which energy is absorbed. There is a question about the final expression for time and its units, as well as the reasonableness of the calculated time.

Discussion Status

Some guidance has been offered regarding the unit cancellation in the final expression. Participants are exploring the implications of the calculated time in relation to practical expectations for heating water in a microwave.

Contextual Notes

There is a potential misunderstanding in the final calculation, as indicated by a participant questioning the reasonableness of the result. The discussion reflects a collaborative effort to clarify the approach without reaching a definitive conclusion.

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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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