Question

My question is too long, so here is a link : http://www.chemicalforums.com/index.php?topic=63499.0 [PLEASE HELP!!!!]

Answer 1

1)
you use:

Q=m cp dT

Q=3125 J = 3,125 kJ
T1= 22°C = 295,15 K
T2= 67,3 °C = 340,45 K
cp = 0,39 kJ / kg K

so

m = Q / cp * dT
m(Zn) = 3,125 kJ / [ 0,39 kJ / kg K * (340,45 K - 295,15 K) ]
m(Zn) = 0,1769 kg

Answer 2

2) again same formula:

Q = m cp dT

but here you have to see that Q gained by the water is Q released by the Ag, so you write:

Q1 = m1 cp1 dT1
Q2 = m2 cp2 dT2

and if:

Q1 = Q2

then:

m1 cp1 dT1 = m2 cp2 dT2

so now you can take that index 1 is Ag and index 2 is water or vice versa

125 g * 0,237 J/g°C * (100 - T2) = 250 g * 4,1813 J/g°C * (T2 -25)

now you solve for T2 and i got that T2 = 27,06 °C

Answer 3

3) same as 2)

Answer 4

4)
you need to calculate in heat of evaporation cause water boils at 100°C which for water is 2257 kJ/kg :)