Alternating Current Problem

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Homework Statement



On its highest setting a heating element on an electric stove is connected to an ac voltage of 240V. This element has a resistance of 29 ohms.(a) Find the power dissipated in the element.(b) Assuming that three-fourths of the heat produced by the element is used to heat a pot of water(the rest being wasted), find the time required to bring 1.9kg of water(half a gallon) at 15degC to a boil

Homework Equations


Vmrs = V0 / square root 2

P=Vrms^2 / R


The Attempt at a Solution



Vrms= 240/sqrt2 = 170V

P= 170^2/29 = 997watts

I'm not sure how to answer part b.

I think I would use the equation:
Q=cm(T1-T2) but I don't know how to solve for time?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
cepheid
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but I don't know how to solve for time?
Solving for Q will give you the total amount of energy needed to heat up the pot by that much. You already know how much useful power the heating element produces. What is the relationship between power and energy?
 
  • #3
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Power is energy per unit time.
 
  • #4
cepheid
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So if you know how much energy the heating element transfers to the pot per unit time, and you know how much total energy it needs, then...
 
  • #5
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so. Q=mc(T-T)
Q= (1.9kg)(4.186)(100-15)
Q=676J

Then Take 3/4 of the P
(997J/s)(3/4) = 748J/s

so it takes 676J to heat the water to boil, and the stove produces 748 J/s towards heating the pot.

676J / 748 J/s = .904 seconds
 
  • #6
Redbelly98
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Homework Statement



On its highest setting a heating element on an electric stove is connected to an ac voltage of 240V. This element has a resistance of 29 ohms.(a) Find the power dissipated in the element.(b) Assuming that three-fourths of the heat produced by the element is used to heat a pot of water(the rest being wasted), find the time required to bring 1.9kg of water(half a gallon) at 15degC to a boil

Homework Equations


Vmrs = V0 / square root 2

P=Vrms^2 / R


The Attempt at a Solution



Vrms= 240/sqrt2 = 170V

P= 170^2/29 = 997watts
Actually, 240V is the rms voltage.

Vrms = 240 V
 
  • #7
rl.bhat
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so. Q=mc(T-T)
Q= (1.9kg)(4.186)(100-15)
Q=676J

Then Take 3/4 of the P
(997J/s)(3/4) = 748J/s

so it takes 676J to heat the water to boil, and the stove produces 748 J/s towards heating the pot.

676J / 748 J/s = .904 seconds
It should be
Q= (1.9kg)(4186)(100-15)
 

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