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Thermal Energy and Heat Transfer

 
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May28-06, 01:57 PM   #1
 

Thermal Energy and Heat Transfer


Question:

How long does it take a 1000W electric kettle to bring 1.0L of water to the boiling point if the initial temperature of the water is 15ºC and the kettle is made of 400g of iron?

Assume that no water is boiled, that no heat is lost to the surroundings, and that the kettle is 100% efficient.

Well first thing I did of course is write givens:
Temperature(water, initial) - 15ºC
Temperature(water, final) - 100ºC
Specific Heat Capacity (water) - 4200 J/kgºC
Specific Heat Capacity (iron) - 460J/kgºC
Mass (iron) - 0.4kg
Mass water - 1.0L = 1kg
Power of kettle - 1000W

I know the p= W/T and e=mct formulas but I'm having trouble with applying the 1000W to the heat transfer formulas. The other questions I've done,only required me to do mcΔt = -mcΔt with some latent fusion/vapourization in there.

Confused where to start...
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May28-06, 02:12 PM   #2
 
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Quote by richievuong
Question:

How long does it take a 1000W electric kettle to bring 1.0L of water to the boiling point if the initial temperature of the water is 15ºC and the kettle is made of 400g of iron?

Assume that no water is boiled, that no heat is lost to the surroundings, and that the kettle is 100% efficient.

Well first thing I did of course is write givens:
Temperature(water, initial) - 15ºC
Temperature(water, final) - 100ºC
Specific Heat Capacity (water) - 4200 J/kgºC
Specific Heat Capacity (iron) - 460J/kgºC
Mass (iron) - 0.4kg
Mass water - 1.0L = 1kg
Power of kettle - 1000W

I know the p= W/T and e=mct formulas but I'm having trouble with applying the 1000W to the heat transfer formulas. The other questions I've done,only required me to do mcΔt = -mcΔt with some latent fusion/vapourization in there.

Confused where to start...
Find how much energy is needed to bring the temperature of the water and iron kettle from 15ºC to 100ºC . Then work out how long it would take to deliver that energy at 1000 Joules/sec.

AM
May28-06, 02:13 PM   #3
 
oh well now i kind of figured that

|E heat lost| = |E heat gained|
so e = mcΔt
e= (1)(4200)(100-15)
e= 357000

p=W/T
1000W = 357000J/T
T = 357s (5.9mins)

answer said 6.2...either i'm missing some steps, or i'm just completely wrong...i didn't fit in the stuff for iron...

EDIT: haha nvm i figured it out
e = mcΔt
e = (0.4)(460)(100-15)
e(iron) = 15640 + 357000 = 372640J/1000= 372.64 (6.2mins)

thanks anyway
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