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sciencegeek101
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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
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