How many watts would it take to raise 50 lbs of water 90 degrees farenheit in 15 minutes?

Question

I got 5274, but I don't know if I did it right. This is for a practical application, not a school problem... Thanks...

Answer 1

Well, this is an interesting question, I'd say you did it right, I came up with an answer of 5.273kW using pounds, degrees F and the value of 3413BTU/kWh (from the table at: http://www.uwsp.edu/CNR/wcee/keep/Mod1/Whatis/energyresourcetables.htm)
There are some slightly different values for BTU so different people might come up with slightly different numbers, but still quite close, except for Norrie, but his (her?) answer would, I think, have been 5.283kW, and quite close to the others, except for making the mistake of converting a temperature of 90°F to degrees C rather than a change of 90°F (you didn't need to subtract 32, since it is only the relative "size" of the degrees we care about, not where we start measuring)

[EDIT] Well, Mr. Norrie, logically your calculation is not irrefutable, because it is wrong. Since some sources give the specific heat capacity of water as joules / gram / degree kelvin (same valeu as if degrees Celsius is used) how about doing your calculation with 90°F converted to kelvin (305.22K) and see what the results are.
I'm pretty sure the person asking the question can convert kilowatts to watts, and anyway I stated out by saying his own result was correct.

[2ND EDIT] Just in case i didn't make myself clear before, consider this situation: The water starts at very close to 32°F (but still liquid), which would be very close to 0°C, then the temperature of the water is raised 90°F to 122°F which is 50°C, and the difference between 0 and 50 is, well, 50, not 32.2. Of course this works with any starting temperature, I used 32°F just because it was easy, but, for example, 68°F (20°C) to 158°F (70°C) gives the same difference

Answer 2

Converting from Imperial to Metric...
(50lb ÷ 2.2lb/kg) x (4.184kJ/kg/°C) x ((90 - 32) ÷ 1.8)°C
= 22.73kg x 4.184kJ/kg/°C x 32.2°C (90°F)
= 3,062 kJ
1 Watt = 1 J/sec.
3,062 kJ ÷ (15min x 60sec/min)
= 3,062kJ ÷ 900s = 3.4 kW (3,400W).

(To 'tinkertailor' guy, 1st, I'm a 'He'. 2nd, for the calculation I performed. which is irrefutable, the °F MUST be converted to °C as the temperature is being RAISED (ΔT) by 32.2°C (90°F) and, I'm using Joules NOT Btu, ..and 1 J/s = 1 Watt...Also, you're looking for Watts not kWh..)

Answer 3

1 BTU = heat necessary to raise 1 lb of water 1 degreeF

so, that's 50*90 = 4500 BTUs of heat

4500 BTU/15min = 5.25 kW

That also assumes the water stays liquid (i.e. less than 212 degF)

.

Answer 4

Watts and watts of them...now be very, very quiet...I am hunting that "pesky wabbitt", He he...shhhhh!