Alternating Current Problem

In summary: JThen take 3/4 of the P, which is 748J/s, therefore 676,670J/748J/s = 904 seconds. In summary, the power dissipated in the heating element on the electric stove is 997 watts and it takes approximately 904 seconds to bring 1.9kg of water at 15 degrees Celsius to a boil using 3/4 of the heat produced by the element. This is calculated by finding the total energy needed to heat up the water using the specific heat capacity and then dividing it by the power produced by the stove, which is 3/4 of the power dissipated in the element.
  • #1
runfast220
25
0

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?
 
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  • #2
runfast220 said:
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
Power is energy per unit time.
 
  • #4
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
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
runfast220 said:

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
runfast220 said:
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)
 

What is alternating current?

Alternating current (AC) is an electrical current that periodically reverses direction. This means the flow of electrons alternates between moving in one direction and then reversing and moving in the opposite direction. It is the most common type of electricity used in homes and businesses.

How does alternating current work?

AC is produced by generators and power plants, where mechanical energy is used to rotate a coil of wire between two magnets. This creates a changing magnetic field, which in turn generates an alternating current in the wire. The frequency or rate at which the current alternates is measured in hertz (Hz).

What are the advantages of alternating current?

One of the main advantages of AC is that it can be easily transformed to different voltages, making it more efficient for long distance transmission. It is also safer for humans to use, as it can be easily controlled and regulated. Additionally, AC allows for the use of transformers, which can increase or decrease the voltage as needed for different applications.

What are the common problems associated with alternating current?

One common problem with AC is voltage drop, which occurs when the voltage decreases as it travels through a wire due to resistance. This can lead to decreased power and inefficiency in electrical systems. Another issue is power surges, which can damage electrical equipment and appliances. It is important to have proper grounding and surge protection to prevent these problems.

How does alternating current compare to direct current?

Direct current (DC) is another type of electricity that flows in only one direction. While AC is more commonly used in homes and businesses, DC is often used in electronics and batteries. The main difference between the two is that AC can be easily transformed and transmitted over long distances, whereas DC cannot. AC is also easier to generate and regulate, making it more practical for widespread use.

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