Easy Thermodynamics Problems....Help!?

Question

hey guys, I need help ASAP! this stuff is due 2morrow, please help!


1. 40.2 g sample of metal is heated to 99.3 C and then placed in a calorimeter containing 120. g of water ( s= 4.184 J/gC) at 21.8 C. the final temp of the water is 24.5 C which medal was used?

b. Iron ( s=0.45 J/gC)
c. Copper ( s=.20 J/gC)
d. Lead ( s=.14 J/gC)
e. Silver (.13 j/gC)


2. consider the following:
2A -> 1/2 B + C change in heat = 5 kJ/mol
5/2 B + 4C -> 2A+ C+ 3D change in heat = -15 kJ/mol
E + 4A -> C

Calculate change in heat for: C-> E + 3D
a. 0 kJ/mol
b. 10
c.-10
d.-20
e20 kJ/mol

E + 4A - > C change in heat is 10 kJ/mol

Answer 1

Total heat remains constant. Before mixing, total heat is Q = 40.2*s*(99.3+273) + 120*4.184*(21.8+273)

Final temp is T is such that

T*40.2*s + T*120*4.184 = Q

fot T = 24.5ºC this becomes

(24.5+273)*[40.2*s + 120*4.184] = Q

calculate Q from above, and solve for s

Answer 2

So read the book and learn to do it on your own if it is easy.