Asked by Sydney
if .38 kJ of heat is added to 19.6 grams of ice at -12.0 degrees celsius, what will be the final temperature?
Answers
Answered by
Joel
Heat gain = Heat loss
0.38kj = mc(T2 - T1)
0.38kj = 19.6g x 2108(T2 - (-12)) (where 2108 is the specific heat capacity of ice.)
0.38 x 1000 = 19.6/1000 x 2108(T2 12)
380 = 0.0196 x 2108(T2 12)
380 = 41.32(T2 12)
380 = 41.32T2 495
380 - 495 = 41.32T2
-116/41.32 = 41.32T2/41.32
-3 = T2 (where T2 = final temp)
0.38kj = mc(T2 - T1)
0.38kj = 19.6g x 2108(T2 - (-12)) (where 2108 is the specific heat capacity of ice.)
0.38 x 1000 = 19.6/1000 x 2108(T2 12)
380 = 0.0196 x 2108(T2 12)
380 = 41.32(T2 12)
380 = 41.32T2 495
380 - 495 = 41.32T2
-116/41.32 = 41.32T2/41.32
-3 = T2 (where T2 = final temp)