Calorimetry A thirsty nurse cools a 2.40 L

[orangem]

Homework Statement

A thirsty nurse cools a 2.40 L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.250 kg and adding 0.123 kg of ice initially at -15.5 degree celsius . If the soft drink and mug are initially at 20.2 degree celsius, what is the final temperature of the system, assuming no heat losses?

Homework Equations

Q1+Q2+Q3+Q4+Q5=0
Q=Mc(change in T)
Q=ML
C ice= 2010 (given)
C water = 4190 (given)
C Al= 904
L = 3.34*10^5 (given)
M drink = Density *Volume = (1000)(.0024)=2.4

The Attempt at a Solution

Q1=M ice (C ice) (Temp) = .123(2010)(0--15.5) = 3832
Q2=M ice (L) = .123 (3.34*10^5) = 41082
Q3=M ice (C water) (Temp) = .123 (4190)(T-0) = 515.4 (T)
Q4=M drink (C water) (Temp) = 2.4 (4190) (T-20.2) = 10056 (T-20.2)
Q5=M Al (C Al) (Temp) = .25(904)(T-20.2) = 226 (T-20.2)

(3832)+(41082)+515.4 (T) + 10056 (T-20.2) + 226 (T-20.2) = 0

I keep getting the wrong answer. I think I am missing a Q or misinterpreting one of them

 

[orangem]

never mind I was right 15.1 is the final temp