What would have been the final temperature of the watere?
2007-11-12 07:05:15 · 5 answers · asked by qwee 1 in Science & Mathematics ➔ Chemistry
100 degrees??????? i dunno know...just trying to understand the question. if not hundred it could be 1,2,3,4,5,6,7,8,9,10,11,12,13... from 0 to infinete...best answer or what...i neads to be from 0-infinite lol!
2007-11-12 07:10:19 · answer #1 · answered by []sunset[]madness[] 2 · 0⤊ 1⤋
specific heat of water is: 4.186 joule/gram °C. The formula relating tempurature to heat in the absence of a phase change is: Q =cm(dt). So:
65x10^3 = 4.186 (450)[Tf - 20] where Tf is the final tempurature. Therefore:
Tf= (65x10^3/[4.186(450)] )+ 20 = 54.5 degrees celcius
2007-11-12 07:27:53 · answer #2 · answered by jeffrcal 7 · 0⤊ 0⤋
q = m x Cg x (Tf - Ti)
q = amount of heat energy gained or lost by substance
m = mass of substance
Cg = specific capacity (J C-1 g-1 or J K-1 g-1)
Tf = final temperature
Ti = initial temperature
Cg for water is 4.18
just plug the numbers into the equation and solve for Tf
2007-11-12 07:23:04 · answer #3 · answered by bustedtaillights 4 · 0⤊ 0⤋
you could shop them in 20 degree C water. 18 is somewhat on the chilly ingredient, basically because of the fact they are extra services to capture parasitic infections, yet they're going to stay at temperature that low. sixty 8-seventy two stages f is right.
2017-01-05 08:41:38 · answer #4 · answered by Anonymous · 0⤊ 0⤋
65,000/(4.186*450) = 34.5°C
34.5°C + 20°C = 54.5°C
2007-11-12 07:19:30 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋