In all chemical equations, regardless of the type of reaction that is occurring, the reagents that are on the left side of the arrow are the reactants and on the right your products.
B. Is the solution.
Heat gained by ice cubes would be equal to the - heat lost by warm water
The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol
Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J
Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C
In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:
q(ice/water) = - q(warm water)
moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)
50.5 g / 18.0 g/mol = 2.81 mol
2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)
16916 + 211.3T2 = -669.4 T2 + 53555
36639 = 880.7 T2
T2 = 41.6 C
(anode +) Cu⁰(s) - 2e⁻ → Cu²⁺(aq)
