If 55 calories of heat is absorbed by 1200 grams of water at 25 celcius what will be the final temperature of the water?
2007-06-08 06:22:28 · 4 answers · asked by pirateprincess 2 in Science & Mathematics ➔ Chemistry
Cp = 75 kJ/mole•deg = 4.2 kJ/g • deg
1 Cal = 4.1867 J
Cp = 4.2*10J/g•deg * (cal/4.1867J) ≈ 1
E = mCp∂T = 1200g (1Cal/g•deg) (x-25deg) = 55 Cal
1200x = 30055
x = 25.046ºC
2007-06-08 06:38:17 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
55/1200 + 25
2007-06-08 13:26:30 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋
1 calorie per gram - degree C.
Delta T = 55/1200 = 0.0458 degrees C
Final temp = 25 + 0.0458 = 25.0458 degrees Celcius
2007-06-08 13:29:31 · answer #3 · answered by Anonymous · 0⤊ 0⤋
If you mean calories with a little c, then the previous answer is correct.
If you mean Calories (nutritional calories, or kilocalories), then multiply 55 X 1000 to get calories, and divide that by 1200 and add to 25.
2007-06-08 13:30:31 · answer #4 · answered by hcbiochem 7 · 0⤊ 0⤋