the temperature of 2.0 kg of water is 100.0 degrees C, but the water is not boiling, because the external pressure acting on the water surface is 3.0 X 10^5 Pa. normally, the boiling point of water at 3.0 X 10^5 Pa is 134 degrees C. determine the amount of heat that must be added to the water to bring it to the point where it just begins to boil?
2007-03-17 19:27:35 · 2 answers · asked by tico 1 in Science & Mathematics ➔ Physics
Specific Heat of water = 1 calorie/ (gram-degree celsius)
Heat Input = Mass X Specific Heat X Change in Temperature
= 2000 grams X 1 X (134-100)
= 2000 X 34
= 68000 calories or 68 kilocalories
If you want the answer in joules 1 calorie = 4.184 joules
=68000 X 4.184
= 284,512 joules
2007-03-17 19:48:52 · answer #1 · answered by geo 1 · 0⤊ 0⤋
Heat = 2*4184*34 = 284.512 kiloJoule
2007-03-18 02:38:08 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋