0.25 kg of ice at -5 degree C is dropped into an insulated 2.9 kg aluminum conta
ID: 1569039 • Letter: 0
Question
0.25 kg of ice at -5 degree C is dropped into an insulated 2.9 kg aluminum container at 65 degree C. After a while, all the ice melts, leaving only water. How much heat energy does the ice gain while it a solid? 2625 J 2526 J 5262 J 6252 J How much heat energy does the ice gain as it melts? 63250 J 73250 J 83250 J 93250 J What is the final temperature of the water? 0 degree C 23 degree C 33 degree C 43 degree C What is the final temperature of the aluminum container? same as in #6 temperature found in #6 plus 4.65 degree C temperature found in #6 plus 42.8 degree C 65 degree CExplanation / Answer
28.
Ti = initial temperature of ice = - 5 C
Tf = final temperature = 0 C
mi = mass of ice = 0.25 kg
ci = specific heat of ice = 2108
heat energy is given as
Q = mi ci (Tf - Ti)
Q = (0.25) (2108) (0 - (-5))
Q = 2625 J
29)
Q' = m L = 0.25 x 334000 = 83250 J
30 )
M = mass of aluminium container
using conservation of heat
Q + Q' + m cw (T - 0) = M Cal (65 - T)
2625 + 83250 + 0.25 (4186) T = 2.9 x 900 (65 - T)
T = 22.91 = 23
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