To help keep his barn warm on cold days, a farmer stores 890 kg of solar-heated water (Lf = 3.35 105 J/kg) in barrels. For how many hours would a 1.0-kW electric space heater have to operate to provide the same amount of heat as the water does, when it cools from 10.0 to 0.0°C and completely freezes?
i am really confused what to do. i thought that the 1.0 kW was work, which equals q/t. so i solved for q by adding the q that it would take to get to zero and the q for fusion. BUT when i went to solve for t i got a negative answer. so i know it must be wrong. and redid it and got a huge number. any advice on how to take this problem?
2007-02-22 10:56:49 · 3 answers · asked by lifewithgooli 1 in Science & Mathematics ➔ Physics
I don't have the numbers for this, but here's what you have to do:
Work out how much heat is needed to raise 890 litres of water from 0 Centrigrade to 10C using the specific heat capacity of the water. (I know this is the wrong way around, but the magnitude of number is the same whether cooling from 10 -> 0 or heating from 0 -> 10). You then need to consider how much energy is released by the freezing of the water as it changes state at 0 centrigrade.
Once you have this number, this is the total of Joules given out from your water in the barrels.
Watts is power, i.e. the rate of energy. So 1kW = 1000 Watts per second.
Divide the number of Joules you have from your answer by 1000 to give you the number of seconds and then divide this number by 3600 (seconds in an hour) to give you the number of hours.
2007-02-22 11:07:31 · answer #1 · answered by davidbgreensmith 4 · 0⤊ 0⤋
You got some good advice in making sure you account for the latent heat of fusion of the water after it freezes completely (breaking all the farmers barrels on him). A Watt is a Joule per second and a KiloWatt is 1000 Watts, (not 1000 Watts/second). A calorie is defined as the amount of energy needed to raise 1 gram (or cc) of water by 1 degree Celius. There is a conversion factor between Joules and Calories, I forget the number, around 4.18 (I think) dont take my word for it, look it up. Then just do the math
2007-02-22 19:39:56 · answer #2 · answered by SteveA8 6 · 0⤊ 0⤋
There are two things you need:
Specific heat of water = 4184 J/(kg-deg)
1 KW = 3.6 10^6 J/hr
Heat of Fusion = 3.35 10^5 J/kg x 890 kg = 298.15 10^6 J
Heat Released on cooling = 4184 J/kg-deg x 10 deg x 890 kg = 37.2376 10^6 J
Total Heat Released = 335.3876 10^6 J (sum of both heats)
Thus
335.3876 10^6 J divided by 3.6 10^6 J/hr = 93.16 hours
2007-02-22 20:03:10 · answer #3 · answered by JJ 1 · 0⤊ 0⤋