Thermodynaics, solving for minimum power to heat water

In summary, an electric hot water heater takes in cold water at 15.8°C and delivers hot water at a constant temperature of 45.6°C with a volume flow rate of 5.0 multiplied by 10-6 m3/s. To calculate the minimum power rating of the hot water heater, the equation Q=cm(deltaT) can be used, where Q is heat energy, c is the specific heat capacity, m is mass, and deltaT is the temperature difference. After converting units, the correct formula for power is Power = 4186(1.94)(5E-6)(45.6-15.8). The density of water should be 1g/ml, or 1g
  • #1
copitlory8
86
0
An electric hot water heater takes in cold water at 15.8°C and delivers hot water. The hot water has a constant temperature of 45.6°C, 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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  • #2
copitlory8 said:
An electric hot water heater takes in cold water at 15.8°C and delivers hot water. The hot water has a constant temperature of 45.6°C, 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?
 
  • #3
Power is just the rate at which energy is converted, remember that.
 
  • #4
Q=cm(deltaT)
 
  • #5
SiYuan said:
Power is just the rate at which energy is converted, remember that.


copitlory8 said:
Q=cm(deltaT)

And we know that mass = density*volume = ρV and then divide by time, what do we get?
 
  • #6
wait I'm getting lost. sorry. so do i do Q=[cpv(deltaT)]/t
 
  • #7
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
 
  • #8
is this correct?
 
  • #9
oh no wait. i got the density wrong. i have no idea what the density is since the temperature is different
 
  • #10
copitlory8 said:
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
 
  • #11
0.591312 m^3?
 
  • #12
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.
 
  • #13
yeah. 1.94 is ft^3

so in the equation i would actually plug in .001kg/m^3 since were using kg not g
 
  • #14
Your unit conversion is wrong, if you were to use m^3 you should get something else.
 
  • #15
Power= 4186(1000)(5E-6)(45.6-15.8)
so the conversion is 1000kg/m^3
is the above formula correct?
 
  • #16
It should be right, yes.
 
  • #17
thank you so much. i actually learned something for once.
 
  • #18
My pleasure, copitlory8
 
  • #19
would it be pushing it if i asked you another physics question?
 
  • #20
That'd be fine, as long as I can answer it.
 
  • #21
u are probably the smartest most helpful online free tutor. every website i go people give me crappy answers to try and seem smart but they never help.

anyways. here it goes.
From a hot reservoir at a temperature of T1, Carnot engine A takes an input heat of 5550 J, delivers 1500 J of work, and rejects heat to a cold reservoir that has a temperature of 495 K. This cold reservoir at 495 K also serves as the hot reservoir for engine B, which uses the rejected heat of the first engine as input heat. Engine B also delivers 1500 J of work, while rejecting heat to an even colder reservoir that has a temperature of T2. Find the temperatures

(a) T1 and
_______K
(b) T2.
_______K
 
  • #22
I'm sorry I can't help you for this.

Im also studying Physics at this moment and the Carnot engine is not in my field of studies. So I suppose your first line is wrong then :)
 
  • #23
okay that's fine. you were still helpful. and if you want to redeem urself here is a slightly easier one:
The drawing shows two thermally insulated tanks. drawing is located at:
http://www.webassign.net/CJ/14_26.gif
They are connected by a valve that is initially closed. Each tank contains neon gas at the pressure, temperature, and volume indicated in the drawing. When the valve is opened, the contents of the two tanks mix, and the pressure becomes constant throughout.

(a) What is the final temperature? Ignore any change in temperature of the tanks themselves.(Hint: The heat gained by the gas in one tank is equal to that lost by the other.)
________K
(b) What is the final pressure?
________Pa
 

1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the study of energy and how it is transferred between different forms. It also studies the relationship between energy, work, and heat.

2. How is thermodynamics related to heating water?

Thermodynamics is directly related to heating water as it deals with the transfer of energy in the form of heat. In the case of heating water, the goal is to increase the temperature of the water by transferring heat energy to it.

3. How can I calculate the minimum power needed to heat water?

To calculate the minimum power needed to heat water, you will need to know the specific heat capacity of water, the mass of the water, and the desired temperature increase. You can then use the equation P = (m * c * ΔT) / t, where P is power, m is mass, c is specific heat capacity, ΔT is temperature change, and t is time.

4. What factors affect the power needed to heat water?

The power needed to heat water can be affected by various factors such as the initial temperature of the water, the mass of the water, the specific heat capacity of water, and the desired final temperature. Other factors such as the type of heating method and the efficiency of the heating system can also play a role.

5. Can I use thermodynamics to determine the most efficient way to heat water?

Yes, thermodynamics can be used to determine the most efficient way to heat water. By understanding the principles of thermodynamics and the factors that affect the power needed to heat water, one can determine the most efficient heating method for their specific situation.

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