A 136.4-g sample of water at 21.1 C is heated to steam at 125.4 C. How much heat was absorbed? ____kcal
2007-02-04 03:26:40 · 3 answers · asked by Hey ;) 3 in Science & Mathematics ➔ Physics
14.227 kcal
RB
2007-02-04 03:33:31 · answer #1 · answered by riobob00 3 · 0⤊ 1⤋
You have to know the heat of vaporization (540 cal/g) and the specific heat of steam (0.48 cal/g-degC).
1. Heating the water from 21.1C to 100C
....= 136.4g * 79.1 degC * 1 cal/g-degC
....= 10,790 cal
2. Vaporization
....= 136.4g * 540 cal/g
....= 73,660 cal
3. Heating the steam from 100C to 125.4C
....= 136.4g * 25.4 degC * 0.48 cal/g-degC
....=1663 cal
Total = 86110 cal
....= 86.11 kCal (to four sig figs)
2007-02-04 11:36:23 · answer #2 · answered by gebobs 6 · 0⤊ 0⤋
125.4 - 21.1 = 104.3 kcal
2007-02-04 11:33:37 · answer #3 · answered by SKG R 6 · 0⤊ 1⤋