How much heat is required to raise the temperature of 22 g of water from 28°C to 80°C?
(80 - 28) x 22 = 1144 calories
Who the FVCK gave this FIRST CORRECT answer the thumbs down??? I'll stick a boot up your dumb ass.
T final=80-28 (m=22)
= 1144 Cal
Mass x Specific Heat x ?T = Joules.
?T = 80 - 28 = 52°C; SH of water = 4.184J/g/°C.
22g x 4.184J/g/°C x 52°C ?T.
= 4,786.5 Joules of heat. (4.787kJ).
(4.787kJ / 4.184kJ/kcal = 1.144kcal).