2007-10-31 04:12:47 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
q = s X m X ∆T
q = 13.9 g X 4.18 J/g ˚C X (95.0 ˚C - 17 ˚C)
q = 4532 J = 4.53 X 10^3 J = 4.53 kJ
2007-10-31 04:23:29 · answer #1 · answered by William Q 5 · 1⤊ 0⤋
you need one calorie or 4.184J to higher temperature of water by one degree
Your foumula is Q= mc (Tf-Ti) where m mass , c specific heat , Tf filal temperature , Ti initial temperature
Q = 13.9 *1*(95-17) =1084.2 calories =4536 J
2007-10-31 04:22:58 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
q = mCpDT
q = 13.9g*(4.1813 J/g K)(95 - 17)
q = 4533 J = 1079 cal = 1.08 kcal
2007-10-31 04:27:37 · answer #3 · answered by landonastar 3 · 0⤊ 0⤋
q = 13.9g x 4.184J/g/°C x 78°C ÎT
= 4,536.3 Joules = 4.536 kJ.
Or...q = 4,536J ÷ 4.184J/cal = 1,084 calories. (1.084kcal).
2007-10-31 06:38:28 · answer #4 · answered by Norrie 7 · 0⤊ 0⤋