Rose Cup

Calculate the heat absorbed (in kJ) by a styrofoam cup calorimeter holding a reaction mixture which rose in?

temperature form 20.17 degrees celcius to 41.5 degrees celcius. (Assume the heat capacity of the cup is 380 J/degrees celcius)

dH = C dT

dH = (380 J / C) (21.33 C rise in temp)

dH = 8105 Joules
rounded to 3 sig figs would be
8110 Joules

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regarding your question at

using these three equations:
(A) 2H2 & O2 –> 2H2O (l) ………….dH = -571.6 kJ
(B) Mg(s) & 2 HCl(aq) –> MgCl2 (aq) & H2 …………… dH = -466.9 kJ
(C) MgO (s) & 2HCl(aq) –> H2O(l) & MgCl2(aq) ……….dH = -151.0 kJ

though Hess’s law:
by using 1 equation (A) you get the O2 that you need
by using 2 equation (B) you get the 2 Mg that you need
by using the reverse of two equation (C)’s you get 2 MgO as product

i.e.
+1 (A) 2H2 & O2 –> 2H2O (l)
+2 (B) 2 Mg(s) & 4 HCl(aq) –> 2 MgCl2 (aq) & 2 H2
- 2 (C) 2 H2O(l) & 2 MgCl2 (aq) –> 2 MgO (s) & 4 HCl(aq)

note that
2 H2O –> cancel out with —> 2 H2O
2 H2 –> cancel out with —> 2 H2
2 MgCl2 –> cancel out with —> 2 MgCl2
4 HCl –> cancel out with –> 4 HCl

leaving you with only what you wished:
2Mg (s) + O2 (g) ———> 2 MgO (s)

so then by hess’s law,
if +1 (A) +2 (B) – 2 (C) gives the equation you wish…
then the energies of +1 (A) +2 (B) – 2 (C) gives the energy you wish…

+1 (A) = (-571.6 kJ)
+2 (B) = 2 (-466.9 kJ) = -933.8 kJ
– 2 (C) = (-2) (-151.0 kJ) = + 302 kJ

which total dH = – 1203.4 kJ

p.s. using the dHf tables at http://bilbo.chm.uri.edu/CHM112/tables/thermtable.htm
for the reaction: 2Mg (s) + O2 (g) ———> 2 MgO (s)
dHf of products minus reactants, gives dH = -1203.4 kJ also

if you email me , I csa address your questions

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p.s.

in regard one of the equations I gave you previously:
MgO (s) & 2HCl(aq) –>2 H2O(l) & MgCl2(l)

I had to make some corrections , they should have been:
(C) MgO (s) & 2HCl(aq) –>1 H2O(l) & MgCl2(aq)

that’s only 1 H2O produced, & then it is “aqueous” MgCl2 produced

2008 Oregon Region SCCA June Portland Rose Cup S2 Regional Race