If you heat 1200 grams of water from 22.3'C to 78.9'C, how many joules of heat did it absorb?
2007-11-30 01:50:18 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Water- 4.184 J/g 'C
2007-11-30 01:50:40 · update #1
Heat=mass x specific heat x change in temperature
heat= 1200 x 4.184 x 56.6
2007-11-30 03:38:56 · answer #1 · answered by science teacher 7 · 0⤊ 0⤋
Q= w * Cp * delta T
w=1200 g
Cp= 4.184 J/g
delta T = 78.9 - 22.3
(you may have another symbol for Cp.. I use the one common in thermodynamics. )
2007-11-30 09:59:51 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
heat absorbed is = mst calories where m=1200grams
s=1(sp.heat of water)
t=(78.9 - 22.3)°C
4.1868 joule = 1 calorie
2007-11-30 10:04:04 · answer #3 · answered by eematters 4 · 0⤊ 0⤋
(4.184 J/g 'C) * (1200g) * (78.9 - 22.3)C = 284.1 kJ
2007-11-30 09:55:36 · answer #4 · answered by LSEaves 2 · 0⤊ 0⤋