# Alternating Current Problem

#### runfast220

1. 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

2. Homework Equations
Vmrs = V0 / square root 2

P=Vrms^2 / R

3. 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?
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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#### cepheid

Staff Emeritus
Gold Member
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?

#### runfast220

Power is energy per unit time.

#### cepheid

Staff Emeritus
Gold Member
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...

#### runfast220

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

#### Redbelly98

Staff Emeritus
Homework Helper
1. 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

2. Homework Equations
Vmrs = V0 / square root 2

P=Vrms^2 / R

3. The Attempt at a Solution

Vrms= 240/sqrt2 = 170V

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

Vrms = 240 V

#### rl.bhat

Homework Helper
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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