Solve Flow Rate Problem: 9 kW Shower

[tweety1234]

Homework Statement

a) A student taking a shower notices that the power rating of the shower is 9 kW. Estimate the flowrate of the water.

I think I have to use the equation Q = m c \delta T

However I have not beeen given any temperature or anything, like that, or the mass? How would I do this question? Is there any other formula to use? Am I just mean to assume an temperature?

Thank you.

 

[Pagan Harpoon]

It says "estimate", so you're probably meant to make reasonable guesses at things like that. Suppose the temperature of the room is 20 degrees, the temperature of the water is maybe 50 degrees. You'll need to make similar guesses for some of the other quantities involved too.

 

[tweety1234]

It says "estimate", so you're probably meant to make reasonable guesses at things like that. Suppose the temperature of the room is 20 degrees, the temperature of the water is maybe 50 degrees. You'll need to make similar guesses for some of the other quantities involved too.

okay,

so Q = 9KW

c = 4.198 KJ kg^{-1} K^{-1}

T = 50C - 20C = 30C

so do I just solve for 'm'? Is that the flow rate?

 

[Pagan Harpoon]

Yes, that is probably the best approach.

I was thinking that you would also account for the fact that some of the shower's power is spent giving the water kinetic energy and you would need to make a guess at the pressure with which it comes out. Now that I look more closely at it, though, I think that effect is probably negligible compared to the heating.

 

[tweety1234]

Hello, Just wondering should I change KW to Watts?

cause when I do the calculation with 9KW I get

9 = m \times 4.198 \times 20

m = 0.086 quite a small flow rate?

Also having a few troubles with the units, if I am calculating flow rate it should be kg/s

but if I do some dimensional analysis on this equation I just get mass ?

I know that W = \frac{J}{s}

so does KW = \frac{kJ}{s} ?

\frac{KJ}{s} = m \frac{KJ}{kg K} K

\frac{KJ}{s} = \frac{m KJ}{kg}

 

[Pagan Harpoon]

I think you did the arithmetic wrong there. Firstly, you're now using 20 degrees for the temperature change? I think that may be too low, but okay.

9/(4.198*20)=0.107

I think this is about right, consider if you held a 1 litre container under a shower, then with this flow rate, it would take 9 seconds to fill. That sound reasonable to me.

As for the dimensions, there is no problem. Consider that the dimensions of m here are kg/s because m is not a mass, it is how much mass passes every second. With this in mind, the equation you wrote balances.