Use 1 BTU = 1 lb of water 1° F.
Quantity = 92 lbs
Temp change = (137° - 119°) = 18° F.
92 x 18 = 1,656 .
It doesn't 'take' any heat to do the job. Since you're allowing the water to cool,
from 137° to 119° , the water will release 1,656 BTU of heat as that happens.
Then you can gather up the heat and take it away to do something else with it.
1 BTU is required to raise 1lb of water 1 degree F in 1 hour. 212-75=137 degrees 600 lbs water x 137 degrees= 82,200 BTU's required to change 75 degree water to 212 degree water. To change 212 degree water to 212 degree steam it requires 970 btu's (latent heat of vaporization) per lb of water 970 btu x 600 lbs water = 582,000 btu Answer - 582,000 btu+ 82,200 btu = 664,200 btu's
135 btu
The idea here is to: * Look up the specific heat of water. * Multiply the mass, times the temperature difference, times the specific heat of water. You may need to do some unit conversions first; specifically, if the specific heat is given per kilogram, you can convert the grams to kilograms.
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
Heat required for this transition is given as the the sum of three heatsheat required for heating the ice from -5 degree Celsius +latent heat(conversion of ice at zero degree to water at zero degrees)+heat required to heat the water from 0 to 5 degree CelsiusHeating of ice=m x s x delta T,where m is the mass ,s is the specific heat of ice=200x0.5x5=500calmelting of ice=mxlatent heat=200x80=16,000calHeating of water=m x s x delta T,where m is the mass ,s is the specific heat of water =200x1x5=1000calTotal heat required=500+16,000+1000=17,500 cal
To change the temperature of water from 27ºC to 32ºC will depend on the mass of water that is present. Obviously, the more water, the more heat it will take. This can be calculated as follows:q = heat = mC∆T where m is the mass of water; C is sp. heat = 4.184 J/g/deg and ∆T is 5ºC (change in temp).
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
Heat required to have such a change of state is called latent heat. If L J/kg is the latent heat per kg of water then for M kg of water we need M* L joule of heat energy
Latent heat is the measurement of energy needed to change the state of a substance at its melting point or boiling point. The latent heat of fusion of water is the amount of energy needed to change a fixed amount of water from a solid to liquid at 0 degrees C. this works out to be more than 800KJ of heat energy. The latent heat of vaporization of water is the amount of energy needed to change a fixed amount of water from a liquid to a gas at 100 degrees C. this is more than 1200KJ of heat needed to be absorbed.
the answer is 200 Btu. the formula is weight x specific heat x temperature difference so we have 10 pounds x 1.00 x 20 10 for 10 pounds of water the specific heat of water is 1.00 and the temperature difference would be 70-50= 20
Then it will either get hotter, or change its phase (for example, ice at zero degrees will convert to water, also at zero degrees).
10 degrees Celsius
1 BTU is required to raise 1lb of water 1 degree F in 1 hour. 212-75=137 degrees 600 lbs water x 137 degrees= 82,200 BTU's required to change 75 degree water to 212 degree water. To change 212 degree water to 212 degree steam it requires 970 btu's (latent heat of vaporization) per lb of water 970 btu x 600 lbs water = 582,000 btu Answer - 582,000 btu+ 82,200 btu = 664,200 btu's
100 degrees celsius are equal to 212 degrees fahrenheit.
Please try to use your brain to solve simple questions such as this one. The change in temperature is simply 100 - 35 = 65 degrees Celsius.
135 btu
The idea here is to: * Look up the specific heat of water. * Multiply the mass, times the temperature difference, times the specific heat of water. You may need to do some unit conversions first; specifically, if the specific heat is given per kilogram, you can convert the grams to kilograms.