it will be 11,700J
since the formula would be:
q=mHf=35g x 334 J/g= 11,700 J
Ice at 0 deg C has a latent heat of melting of 335 kJ/kg. If you have 35 grams (0.035 kg), then it would take exactly 11.725 kJ to melt the ice to get 35 grams of water at 0 deg C.
42 , and 28.
43.2 grams of water
The temperature would be that of water's boilng point od 100 degrees
66g
That completely depends on the specific heat capacity of the substance of which the sample is composed, which you haven't identified. It will take a lot more heat energy to raise the temperature of 65 grams of water 35 degrees than it would take to do the same to 65 grams of iron or yogurt, e.g.
Ice at 0 deg C has a latent heat of melting of 335 kJ/kg. If you have 35 grams (0.035 kg), then it would take exactly 11.725 kJ to melt the ice to get 35 grams of water at 0 deg C.
Change in temp = 28° - 22° = 6°. Heat capacity of liquid water is 1 calorie per gram per degree(C).So the energy absorbed is (50) x (6) = 300 calories.
158.1Kj
Grams solid × mol/g × Hfusion
Grams liquid × mol/g × Hvap
Grams liquid × mol/g × Hvap
(75'C)x(1g) < (75'C)x(100g) .'. The second option has more thermal energy.
q(joules) = mass * specific heat * change in temperature ( 8 kg = 8000 grams ) q = (8000 grams H2O)(4.180 J/gC)(70o C - 20o C) = 1.7 X 106 joules ============
21 grams through 71 degrees is 21x71 calories.
k
Exactly 5 grams of antimatter will completely annihilate 5 grams of matter, producing an enormous shower of high energy gamma rays. In total a mass of 10 grams will be converted to the equivalent amount of energy.