How many watts are required to raise 1 gallon of water 1 degree Fahrenheit?

Answer 1

To raise 1 gallon of water by 1 degree Fahrenheit, you would need approximately 8.33 BTUs (British Thermal Units) of energy. One BTU is equivalent to 0.29307107 watts, so to convert this to watts, you would need about 2.44 watts of power. This calculation assumes ideal conditions and does not account for any energy losses during the heating process.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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.

Answer 7

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

Answer 8

This cannot be answered without knowing the elevation (distance above sea level) at which the water is being heated.

Answer 9

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.

Answer 10

498,000 BTU