please show all work I am confused on how to get it. 54.0 g aluminum at 50.0 o C
ID: 102674 • Letter: P
Question
please show all work I am confused on how to get it.
54.0 g aluminum at 50.0oC is dumped into water at 38.5oC. The water warms to 40.0oC. (10 pt)
What is specific heat capacity of aluminum?
a) ____________
What is molar heat capacity of aluminum?
b) ____________
When 10.0 g ammonium nitrate is dissolved in 50.0 g water at 25.0oC, the final temperature of the solution is 20.0oC.
a. What is the q of the water? (SH water = 4.20 J/g K) (10 pt)
qH2O _________
b. What is the molar change in enthalpy (H in kJ/mol NH4NO3) for the solution process?
H ___________
Explanation / Answer
heat lost by Al = heat gained by water
mass of Al*SAl*DT = mass of water*S*DT
54*s*(50-40) = mass of water*4.184*(40-38.5)
mass of water not given
2)
a) q of the water = m*s*DT
= (50)*4.2*(25-20)
= 1050 joule
qH2O = 1050 joule
b. no of mol of Al(NO3)3 dissolved = 10/213 = 0.047 mol
molar change in enthalpy (H) = -q/n
= -1.05/0.047
DHrxn = -22.3 kj/mol
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