When a piece of Aluminum weighing 35.7 g, and at a temperature of 81.9 degrees C, is placed in a calorimeter containing 75.0 g of water at 24.9 degrees C, the temperature increases to 28.3 degrees C. If the specific heat of water is 4.184 J/gK and the specific heat of Aluminum is .0902 J/gK, what is the HEAT CAPACITY of the CALORIMETER?
2006-11-26 14:20:02 · 3 answers · asked by Captain Whiskerboy Litterbox 3 in Science & Mathematics ➔ Chemistry
The change in heat = specific heat * mass * change in temperature (Q=C*g*ΔT).
Assuming that the Aluminum also ends at 28.3 °C, then the change in heat for the Aluminum is
0.0902 J/g°K * 35.7 g * (28.3°-81.9°)
The above will be negative as heat was lost from the Aluminum. The change in heat for the water is
4.184 J/g°K * 75.0 g * (28.3°-24.9°)
The above will be positive as the water gained heat.
The Heat Capacity of the calorimeter is the difference between the heat lost by the Aluminum and that gained by the water divided by the change in temperature (which is the same as for the water (28.3°-24.9°).
2006-11-27 09:41:19 · answer #1 · answered by Richard 7 · 66⤊ 0⤋
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2016-12-29 13:07:53 · answer #2 · answered by Anonymous · 0⤊ 0⤋
194 J/K
2013-11-03 08:44:14 · answer #3 · answered by Mo 2 · 1⤊ 0⤋