Asked by Ish

A 28.4 g sample of aluminum is heated to 39.4 C, then is placed in a calorimeter containing 50.0 g of water. Temperature of water increases from 21.00 C to 23.00 C. What is the specific heat of aluminum in cal/gC?
Is the answer for this 0.3, using sig figs? Do you find the q of water and then set it equal to 28.4 (amt of grams) * C (The specific heat) * 11 (which is the temp difference)
Answered by DrBob222
Close but I don't know where you came up with 11 difference. The specific heat Al is about 0.9 J/g*C. You can put all of it into one equation as follows:
[(mass H2O x sph.H2O x (Tf-Ti)] + [(mass Al x sph.Al x (Tf-Ti)] = 0
[50*4.184*(23.00-21.00)] + [28.4*x*(23-39.4)] = 0
Solve for ? and I get approx 0.898
Answered by Anonymous
That is wrong.
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