Physics Homework about using Electricity to Heat Flowing Water

In summary, you need to calculate the power given out by the resistor in order to heat the water. The resistance will determine how much power is needed.
  • #1
Detopall
2
0
New user has been reminded to show their work on schoolwork questions
Homework Statement
This is a question of my Physics homework: Electricity
Relevant Equations
Q = c m ΔƟ
I = Q/t
R= U/I
This is the question: You want to make an electric instantaneous water heater in which 5.0 liters of water flows past a resistance per minute and heats water from 10.0 ° C to 45.0 ° C. Calculate the magnitude of the resistance to use and the amperage. The flow-through is connected to 230V

So: Given: 1min = 60sec = t// 5.0l = volume of water= 5kg = m // ΔT = 35°C // V = 230V //
Asked: R & I
Answer: Q = c . m . ΔT
Q = 4186 . 5 . 35
Q = 732550 J/kg . K
I = Q/ t
I = 732550/ 60
I = 12209.17 A
 
Last edited:
Physics news on Phys.org
  • #2
Hi @Detopall,

:welcome:

What have you tried so far to solve the problem?
 
  • #3
Your "relevant equations" are all electrical. What about the heat?
 
  • #4
Merlin3189 said:
Your "relevant equations" are all electrical. What about the heat?
Detopall said:
Q = c m ΔƟ
This looks pretty much like heat relation to temperature through mass and specific heat capacity. Using Q to denote both heat and charge is however ... questionable ... at best.
 
  • Like
Likes Merlin3189 and berkeman
  • #5
Orodruin said:
This looks pretty much like heat relation to temperature through mass and specific heat capacity. Using Q to denote both heat and charge is however ... questionable ... at best.
Thanks @Orodruin -- for the life of me, I couldn't decode that first equation (but I was starting with the "electrical" Q)... o0)
 
  • #6
Yeah, right. Sorry. Like berkeman I just saw the Q's and thought, something about charge, irrelevant.
It's not a formula I've used for a long time (if ever?), whatever letters. I just remember it as the mechanical equivalent of heat is 4.2 J/cal. I never really think about water as having a specific heat capacity until it comes to the latent heat.
 
  • Like
Likes berkeman
  • #7
So: Given: 1min = 60sec = t// 5.0l = volume of water= 5kg = m // ΔT = 35°C // V = 230V //
Asked: R & I
Answer: Q = c . m . ΔT
Q = 4186 . 5 . 35
Q = 732550 J/kg . K
I = Q/ t
I = 732550/ 60
I = 12209.17 A

I don't even need to continue. 10 mA is deadly.

I don't know what I did wrong. Can anybody help? Thanks.
 
  • #8
Detopall said:
I don't know what I did wrong. Can anybody help? Thanks.
Orodruin said:
Using Q to denote both heat and charge is however ... questionable ... at best.
... and more likely leads to serious errors when you just blindly insert heat instead of charge into I = Q/t.
 
  • Haha
  • Like
Likes berkeman and Merlin3189
  • #9
Detopall said:
Q = 732550 J/kg . K
In addition to the above, this is also wrong. The units of heat is Joules. You have given the units of the specific heat ##c##.
 
  • #10
Detopall said:
I don't know what I did wrong. Can anybody help? Thanks.
But do you know what you did? (and understand it maybe?)

Little in your working shows what you are finding in the context of the question. You are substituting numbers into algebraic expressions and doing arithmetic calculations perfectly well. But why? What does it all mean? Don't use formulae (equations) blindly. Think what they mean.

The water heater needs to give energy to the water to heat it up.
So you've worked ot the amount of energy needed to heat 5l of water by 35 K. (You should sort out the units here as Orudruin says.)

That energy has to come from the resistor.

As Orudruin says, your formula ##I=\frac Q t ## does not clculate energy. It says current is charge / time

Do you know how to work out the power or energy given out by a resistor in an electric circuit?BTW "10 mA is deadly." I believe the deadly figure is over 30 mA. That's when ELCB's trip. But don't worry, you're going to need a lot more than that to heat this water. Just don't put your finger on the wires when it's switched on:smile:
 

1. How does electricity heat flowing water?

Electricity heats flowing water through a process called Joule heating. When an electric current passes through a conductor, such as a heating element in a water heater, it encounters resistance. This resistance causes the electric energy to be converted into heat energy, which then raises the temperature of the water.

2. What factors affect the efficiency of using electricity to heat flowing water?

The efficiency of using electricity to heat flowing water can be affected by several factors, including the type and size of the heating element, the temperature of the water, and the insulation and design of the water heater. Other factors such as the quality of the electrical connection and the voltage can also impact efficiency.

3. Can electricity be used to heat water instantly?

Yes, electricity can be used to heat water instantly. This is commonly seen in electric kettles, where the heating element quickly heats up the water as it passes through. However, for larger volumes of water, it may take longer for the water to reach the desired temperature.

4. Is using electricity to heat flowing water more efficient than using other methods?

It depends on the specific circumstances and the efficiency of the equipment being used. In general, using electricity to heat flowing water can be more efficient than other methods such as gas or oil heating, as there is no energy loss through flue gases. However, the cost of electricity and the efficiency of the heating element can also impact the overall efficiency.

5. Can electricity be used to heat water in a sustainable way?

Yes, electricity can be used to heat water in a sustainable way. This can be achieved through the use of renewable energy sources, such as solar or wind power, to generate the electricity. Additionally, energy-efficient water heaters and proper insulation can also contribute to a more sustainable use of electricity for heating water.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top