7.014*10^4
one trillion pesos
The specific heat of the substance.
(110 calories) plus (heat lost from the container during the procedure)
1270.
6 % of 1000 = 6/100 * 1000 = 0.06 * 1000 = 60
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
Can you help
314j
19.7 kJ
46389000 j
You need to add all of the following:* The heat required to heat ice from -5 to 0 degrees. Multiply the mass times the temperature difference times the specific heat of ice. * The heat required to melt ice. Multiply the mass by the heat of fusion. * The heat required to raiste the temperature of water from 0 to 20 degrees. Multiply the mass times the temperature difference times the specific heat of water.
1,832 degrees Fahrenheit.
Assuming standard atmospheric pressure, 2260 kilojoules.
If you warm it from 35 degrees Celsius to 1000 degrees Celsius, a mas will vastly increase in volume or pressure. Without knowing how you intend to allow for that, your question is unanswerable.
1000 sq ft
a lot
q = mass * specific heat * change in temperature 1 kg silver = 1000 grams q = (1000 g Ag)(0.237 J/gC)[(- 40o C) - (- 20o C)] = - 4.7 X 103 Joules =============