How Hot Was the Campfire's Aluminum Can Before Cooling in Water?

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In attempt to determine the temperature of his campfire, a teacher removes 15 g aluminum pop can (Caluminium = 900 J/kg•°C) that has been sitting in the coals from the fire and places it into a bucket that contains 2.0 L of water (Cwater = 4 200 J/kg•°C) at 18 °C; 2.0 L of water has a mass of 2.0 kg. If the final temperate of the water, with the can in it, is 19 °C, what was the temperature of the aluminum can when it was in the fire?

I got an answer of -622.22222222222 which doesn't make sense because the can would b hot because its in a fire... could someone please help?

My answer:
m(water)= 2kg
m(aluminum)= 0.015kg
C(aluminum)=900J/kg degrees Celsius
C(water)= 4200J/Kg degrees Celsius
T(water)= 19 degrees Celsius - 18 degrees Celsius= 1 degrees Celsius
T(aluminum)= ?

Q= heat energy lost or gained
m= mass
T= change in temperature
c= heat capacity of the substance

Q(water)= mct
=2(4200)(1)
=8400

Q(water)= -Q(aluminum)
Q(aluminum)= -8400

T(aluminum)= -8400/0.015(900)
= -622.2222222
 
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Assuming that your arithmetic is correct then you are saying that the can lost 8400 Joules of heat energy which would then mean that it lost 622 degrees Celsius, NOT that its initial temperature was -622 degrees.
 
It's a careless treatment of signs. You got -8400J for the change in thermal energy of the aluminum can. This is certainly correct as it means that the aluminum can lost thermal energy to the water when it was placed in the water. -622.2 hence refers to the change in temperature of the can after it was placed in the water.