Want this question answered?
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
question makes no sense.....
Assuming standard atmospheric pressure, 2260 kilojoules.
That completely depends on the temperature of the water when the gas flame is first ignited, which you've neglected to mention.
You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.
28lbs
28lbs, about as much as my 10 month old son.
heat of fusion
How much heat (in calories) is required to heat a 43 g sample of aluminum from 72 F to 145F
how much heat is required to convert 0,3kg of ice at 0c to water at the same tempture
How much heat energy is required to raise the temperature of 0.358 of copper from 23.0 to 60.0 ? The specific heat of copper is 0.0920
125.6kj (apex)
lf = 3.35 x 105 J kg-1 This much amount of heat required to convert 1 kg of ice to liquid Mani.Ra
Heat required = mass x specific heat of water x temperature difference Here we have heat required = 21 x 1 x 10 = 210 cals
The values are different for each type of plastic.
it depends how cold the ice is
No heat (energy) is required to freeze water (from liquid to solid). Freezing RELEASES energy (heat), as it is an exothermic event. If you want to know how much energy is release, you need to know the heat of fusion for water, and then multiply that by the mass of water being frozen.