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Thermodynaics, solving for minimum power to heat water

by copitlory8
Tags: heat, minimum, power, solving, thermodynaics, water
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copitlory8
#1
Sep19-10, 02:17 PM
P: 86
An electric hot water heater takes in cold water at 15.8C and delivers hot water. The hot water has a constant temperature of 45.6C, when the "hot" faucet is left open all the time and the volume flow rate is 5.0 multiplied by 10-6 m3/s. What is the minimum power rating of the hot water heater?

I really need help. I don't know how to connect Power and Thermal Energy. I know the individual formulas but can't connect the two. This is for WebAssign and is due within the hour. Please provide quick solution. Thank you very much.
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rock.freak667
#2
Sep19-10, 02:19 PM
HW Helper
P: 6,208
Quote Quote by copitlory8 View Post
An electric hot water heater takes in cold water at 15.8C and delivers hot water. The hot water has a constant temperature of 45.6C, when the "hot" faucet is left open all the time and the volume flow rate is 5.0 multiplied by 10-6 m3/s. What is the minimum power rating of the hot water heater?

I really need help. I don't know how to connect Power and Thermal Energy. I know the individual formulas but can't connect the two. This is for WebAssign and is due within the hour. Please provide quick solution. Thank you very much.
Ok well let's start simple, what equation connects a temperature difference and heat energy?
SiYuan
#3
Sep19-10, 02:21 PM
P: 27
Power is just the rate at which energy is converted, remember that.

copitlory8
#4
Sep19-10, 02:23 PM
P: 86
Thermodynaics, solving for minimum power to heat water

Q=cm(deltaT)
rock.freak667
#5
Sep19-10, 02:28 PM
HW Helper
P: 6,208
Quote Quote by SiYuan View Post
Power is just the rate at which energy is converted, remember that.

Quote Quote by copitlory8 View Post
Q=cm(deltaT)
And we know that mass = density*volume = ρV and then divide by time, what do we get?
copitlory8
#6
Sep19-10, 02:41 PM
P: 86
wait i'm getting lost. sorry. so do i do Q=[cpv(deltaT)]/t
copitlory8
#7
Sep19-10, 02:46 PM
P: 86
so for my answer it would be Power= 4186(1.94)(5E-6)(45.6-15.8) where c=4186 and the density of water is 1.94
copitlory8
#8
Sep19-10, 02:46 PM
P: 86
is this correct?
copitlory8
#9
Sep19-10, 02:48 PM
P: 86
oh no wait. i got the density wrong. i have no idea what the density is since the temperature is different
SiYuan
#10
Sep19-10, 02:53 PM
P: 27
Quote Quote by copitlory8 View Post
so for my answer it would be Power= 4186(1.94)(5E-6)(45.6-15.8) where c=4186 and the density of water is 1.94
The density of water is 1.94 for feet^3, while your question is in m^3
copitlory8
#11
Sep19-10, 03:11 PM
P: 86
0.591312 m^3?
SiYuan
#12
Sep19-10, 03:13 PM
P: 27
The average density of water should be 1g/ml, or 1g/cm^3,

whilst 1.94 is for lb/ft^3 if I recall right.
copitlory8
#13
Sep19-10, 03:23 PM
P: 86
yeah. 1.94 is ft^3

so in the equation i would actually plug in .001kg/m^3 since were using kg not g
SiYuan
#14
Sep19-10, 03:24 PM
P: 27
Your unit conversion is wrong, if you were to use m^3 you should get something else.
copitlory8
#15
Sep19-10, 03:28 PM
P: 86
Power= 4186(1000)(5E-6)(45.6-15.8)
so the conversion is 1000kg/m^3
is the above formula correct?
SiYuan
#16
Sep19-10, 03:31 PM
P: 27
It should be right, yes.
copitlory8
#17
Sep19-10, 03:35 PM
P: 86
thank you so much. i actually learned something for once.
SiYuan
#18
Sep19-10, 03:35 PM
P: 27
My pleasure, copitlory8


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