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

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SUMMARY

The discussion focuses on calculating the time required to boil water in a microwave oven, specifically using a microwave power output of 1219 W and a water absorption rate of 64.4%. The heat required to raise 212 grams of water from 19.6°C to 100°C is calculated as 71349.53 Joules. The rate of energy absorption is determined to be 785 Joules per second, leading to an incorrect conclusion that the time required is 0.011 seconds. Participants in the discussion highlight the need to reassess the calculation, particularly regarding unit cancellation and the reasonableness of the result.

PREREQUISITES
  • Understanding of the specific heat capacity formula: Q = mcΔT
  • Familiarity with power calculations: P = dW/dt
  • Basic knowledge of unit conversion and dimensional analysis
  • Concept of energy absorption in microwave ovens
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  • Review the principles of energy transfer in microwave heating
  • Learn about unit conversion and dimensional analysis in physics problems
  • Explore the specific heat capacity of different substances
  • Investigate common misconceptions in thermal energy calculations
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Students studying thermodynamics, physics educators, and anyone interested in practical applications of energy transfer in cooking methods.

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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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