Here are some tools you'll need. 1 watt = 1 joule per second 1 joule = 0.239 calorie 1 gallon of water weighs 8.34 pounds 1 pound = 453.6 grams 180 degrees F = 82.2 degrees C 250 degrees F = 121.1 degrees C 1 minute = 60 seconds Remember that a calorie is the amount of heat (energy) required to raise one gram of water one degree C. So, first determine the amount of heat energy required to heat the quantity of water you have. That calculation is independent of time. Then divide by the time to determine the power (in watts) required. I'm thinking your answer should be in the range of a thousand watts or so. One confounder may be the upper temperature, which is well above the boiling point of water at standard pressure. I'm not sure how that affects the computation. Perhaps your teacher didn't select his upper temperature carefully.
by using the formula "Q=mc∆@" where; Q=energy required(J) m=mass of water(1 gallon change to kg). c=specific heat capacity of water(4200). ∆@=temperature change(100-1; if u want to raise the temperature up to 100). eg. Q=100x4200x99 Q=answer
1 US gallon = 3.8 liters = 3.8 kg water = 8.36 pounds, 100 gallons = 836 pounds. So if your degree is Fahrenheit, the energy is 836 BTU. If it is Centigrade this will be 1505 BTU.
Specific heat capacity
Q = cmT
You need to formulate the equation (and convert it to Si) to make it easy to solve.
Temp. Change = 49 - 21 = 28 degrees C
Volume of water = 379 liters
The specific heat of water at 35 degrees Celsius (21 + .5*28) is about 4.178 kJ/kgK
The density of water around your range is about 994 kg/m3
There are 3.6 kilojoules per Watt hour (a joule is watts X seconds)
Q = cmT Where:
Q = heat added in kilojoules
c = specific heat in kilojoules/kilogram degrees Celsius
m = mass in kilograms
T = temperature change So:
Q = 4.178 * (379 * .994) * 28
Q = 44070.9144 kilojoules
44070.9144 kilojoules / 3.6 kilojoules per watt hour = 12241.9207 watt hours
If you have a 1000 watt heater it would take:
12241.92 whr/ 1000 watt = 12.24 hours
To raise a gallon of water by 1 degree F needs a particular amount of energy, which is power times time.
A gallon is 4540 grams, and to raise it by 1 degree F or 5/9 of a degree C would need 4540 x 5/9 calories, or 2522 calories, equal to 10593 Joules (at 4.2 Joules per calorie). So the energy is 10593 Joules, equal to 10593 watt-seconds, therefore 100 watts for 105.93 seconds would do the trick.
The heat capacity of water is 1 calorie / g / °C. But you multiply heat capacity times mass times the change in temperature, so you need to know the starting temperature.
If you have liquid water at 0°C then the change is 100°, so in that case it would be 6000 calories.
The specific heat for water is 4.184 J/g/deg or 1 cal/g/degq = heat = mC∆T
3785 mls = 1 gallon and assuming density = 1 g/ml, 1 gallon = 3785 g = 3.785 kg
3.785 kg/gal x 1000 gal = 3785 kg = mass of water
q = 3785 kg x 4.184 kJ/kg/deg x 25 deg = 395,911 kJ of energy needed
This cannot be answered without knowing the elevation (distance above sea level) at which the water is being heated.
When 1,000 gallons of water is cooled from 160F to 120F, approximately 34.5335657 kJ/kg-C is liberated. This is the top level of the heat capacity available.
498,000 BTU
100
what is the answer to this
Assume the water is initially at 40 F and 14.7 psia.m = ( 1.0 gal ) ( 1.0 ft^3 /7.4805 gal ) ( 62.43 lbm / ft^3 ) = 8.346 lbmQ = Delta U = ( m ) ( Cv ) ( T2 - T1 )Q = ( 8.346 lbm ) ( 1.00 Btu / lbm - Fdeg ) ( 200 F - 40 F ) = 1335 Btu
specific heat
1 calorie is the energy required to raise 1 gram of water by 1 degree C. So it would take 5 calories to raise it by 5 degrees C.
Answer:A BTU is the amount of heat energy required to raise 1 pound of water by 1 degree Fahrenheit. So a 30,000 BTU burner can heat 30,000 pounds of water by 1 degree Fahrenheit. Any multiple of the amount of water and the temperature rise that yields 30,000 is equivalent.And 30 000 BTU equal 31 651,68 kilojoules.Usually it is more meaningful to rate a heating device in BTU/hour.
what is the answer to this
This is a pretty straightforward calculation. By definition, a BTU is the amount of energy required to raise one pound of water one degree F. But you have one gallon of water, which weighs approximately* 8.34 pounds. So, you'd need 8.34 BTU to increase one gallon of water one degree F. Note how the amount of time was not important. Whether you heat the water slowly or quickly doesn't matter. You will still require 8.34 BTU to raise the temperature of a gallon of water one degree F. * I say approximately because the weight of water varies slightly with its temperature. Water is at its densest at 4 degrees Celsius (39 degrees F). A gallon of water at temperatures above and below that value will weigh less.
The British Thermal Unit (BTU): The Amount of work required to raise one pound of water 1 degree Fahrenheit.
British Thermal Unit, the amount of energy required to raise the temperature of one pound of water one degree Fahrenheit
Depends on how high you want to raise the gram of water ;).
Heat capacity is the amount of heat energy required to raise the temperature of a unit mass of a material one degree.
Assume the water is initially at 40 F and 14.7 psia.m = ( 1.0 gal ) ( 1.0 ft^3 /7.4805 gal ) ( 62.43 lbm / ft^3 ) = 8.346 lbmQ = Delta U = ( m ) ( Cv ) ( T2 - T1 )Q = ( 8.346 lbm ) ( 1.00 Btu / lbm - Fdeg ) ( 200 F - 40 F ) = 1335 Btu
It takes 1 BTU to raise 1 lb of water per degree Fahrenheit.
Energy required to raise 1 gramme of water by 1 degree C = 1 calorie also, 1 calorie = 4.186 Joules
Finish your Question. 1 Watt is the amount of energy required to raise 1 liter of water by 1 degree----I think.
It takes 8.33 BTU to raise the temperature of water 1 degree F.
195 joule..