Question

500 ml of 1.0 M NaOH(aq) is added to 500 ml of 1.0 M HCl(aq) and the solution is quickly stirred. The rise in temperature (∆T1) is measured. The experiment is repeated using 100 cm3 of each solution and the rise in temperature (∆T2) is measured. (assume 4.18 J/g-oC and 1.00g/mL) It is found that: (a) ∆ T2 = 5∆T1; (b) ∆T1 = 5∆ T2; (c) ∆ T1 = ∆ T2; (d) ∆ T1 = 4∆ T2; (e) need to know the enthalpy of reaction.

Why is it c?

Answer 1

Theoretical concept:

Temperature is an INTENSIVE property, meaning it does NOT depends on mass... that should tell you something about the answer.

Anyways

mathematically:

Q1 = m1*cp*(Tf1-Ti1)

Q2 = m2*cp*(Tf2-Ti2)

note that in reallity:

Q1/m1 = Cp*(Tf1-Ti1)

Q2/m2 = Cp(Tf2-Ti2)

Cp for both reactions is the same (its water, cp = 4.184)

so we can send it also to the right

Q1/(m1*cp) = (Tf1-Ti1)

Q2/(m2*cp) = (Tf2-Ti2)

Now, since Q depends directly on mass, the less mol you use, the less heat (proportionally since it is the same solution)

therefore

Q1/(m1*cp) = C1

Q2/(m2*cp) = C2

note that experimentally, you will find out that C1 = C2 since this is constant for every amount of concnetrations of NAOH and HCl (i.e. they have the same heat of reaction PER unit mol)

therefore

C = (Tf1-Ti1)

C = (Tf2-Ti2)

then

(Tf1-Ti1)= (Tf2-Ti2)

which is

dT1 = dT2

answer C