Physics Homework about using Electricity to Heat Flowing Water

[Detopall]

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

 

[etotheipi]

Hi @Detopall,



What have you tried so far to solve the problem?

 

[Merlin3189]

Your "relevant equations" are all electrical. What about the heat?

 

[Orodruin]

Your "relevant equations" are all electrical. What about the heat?

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.

 

[berkeman]

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)...

 

[Merlin3189]

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.

 

[Detopall]

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.

 

[Orodruin]

I don't know what I did wrong. Can anybody help? Thanks.

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.

 

[Orodruin]

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 $c$.

 

[Merlin3189]

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