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 calculate the heat required to cool steam to water at a lower temperature, you can use the formula: Q = mcΔT Where: Q = heat energy m = mass c = specific heat capacity ΔT = change in temperature Given: m = 5 pounds ΔT = (232 - 162) = 70 degrees F Specific heat capacity of water = 1 cal/g°C Convert pounds to grams: 1 pound ≈ 453.592 grams Now, plug in the values into the formula to calculate the heat energy.
To change 10 pounds of ice at 20 degrees Fahrenheit to steam at 220 degrees Fahrenheit, you need to supply enough energy to first melt the ice, then heat the water to the boiling point, and finally convert it to steam. This process requires approximately 180 BTUs per pound of ice to melt it, 180 BTUs per pound of water to heat it to the boiling point, and then 970 BTUs per pound of water to convert it to steam. So, for 10 pounds of ice, the total BTUs required would be around 18,300 BTUs.
To change the temperature of water by one degree Fahrenheit, 1 Btu of heat must be added or removed from 1 pound of water.
To change the state of water from solid (ice) to liquid (water), heat must be added to increase the temperature above 0 degrees Celsius. To change water from liquid to gas (water vapor), heat must be added to increase the temperature above 100 degrees Celsius.
To calculate the energy needed to change ice at -32.9 degrees to water at 75 degrees, you need to consider the energy required for three steps: Heating ice from -32.9 degrees to 0 degrees (specific heat capacity of ice) Melting ice at 0 degrees into water at 0 degrees (latent heat of fusion of ice) Heating water from 0 degrees to 75 degrees (specific heat capacity of water) Once you have the energy needed for each step, you can add them together to find the total energy required.
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 enthalpy change for converting 1 mol of ice at -50 degrees Celsius to water at 70 degrees Celsius is the sum of the enthalpy changes for the following processes: heating ice from -50 degrees Celsius to 0 degrees Celsius (heat of fusion), melting ice at 0 degrees Celsius, and heating water from 0 degrees Celsius to 70 degrees Celsius (specific heat of water).
False. When water reaches 212 degrees Fahrenheit (100 degrees Celsius) and boils, it is absorbing both sensible heat (temperature increase) and latent heat (phase change from liquid to gas).
The heat lost by water at 0 degrees Celsius to change to ice is equal to the heat of fusion of water, which is about 334 joules per gram. So, for 2 grams of water, the heat loss would be 2 * 334 = 668 joules.
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.
Then it will either get hotter, or change its phase (for example, ice at zero degrees will convert to water, also at zero degrees).