73.5 g of aluminum is heated in boiling water to a temperature of 98.7°C. ... (rest of question in description)?
73.5 g of aluminum is heated in boiling water to a temperature of 98.7°C. the Aluminum is then placed in a calorimeter containing 1500.0 g of water at a temperature of 25.40°C. The temperature of the water in the calorimeter increase to a final temperature of 28.20°C. What is the specific heat of the Aluminum?
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If you assume that the calorimeter is ideal and absorbs no heat, then the heat that the water absorbed is:
q = m c (T2-T1)
q = 1500.0 g (4.184 J/gC) (28.20 - 25.40 C) = 1.757X0^4 J
So, the aluminum lost -1.757X10^4 J. So for the aluminum:
q = m c (T2- T1)
-1.757X10^4 J = 73.5 g (c) (28.20 - 98.7 C)
c = 3.39 J/gC
Check your numbers in the problem. This calculated value is very far from the actual specific heat of Al.