Quick Chem question

[thedudeo]

I was doing a QB question, but I don't understand this one.

 

"When 100cm^3 of 1.0 mol dm^-3 HCl is mixed with 100cm^3 of 1.0 mol dm^-3 NaOH, the temperature of the resulting solution increases by 5.0C. What will be the temperature change, in C, when 50 cm^3 of these two solutions are mixed?"

A) 2.5

B) 5.0

C) 10

D) 20

 

The answer given was B, but I don't understand why. I was hoping someone here could explain it to me.

Thanks in advance

[sameera95]

I was doing a QB question, but I don't understand this one.

 

"When 100cm^3 of 1.0 mol dm^-3 HCl is mixed with 100cm^3 of 1.0 mol dm^-3 NaOH, the temperature of the resulting solution increases by 5.0C. What will be the temperature change, in C, when 50 cm^3 of these two solutions are mixed?"

A) 2.5

B) 5.0

C) 10

D) 20

 

The answer given was B, but I don't understand why. I was hoping someone here could explain it to me.

Thanks in advance

because they're the still the same concentration and of equal volumes, therefore equal molar concentrations (1:1). so the outcome will be the same, despite there being less of each. It's more about the molar ratio than the volume I hope it makes sense

[Ossih]

Think about it this way 

 

Q = mc∆T

So, ∆T = Q/(mc)

Now, because the volume of each halves, the mass halves so ∆T goes up by 2. But, because the concentration is the same and the volume halves, the number of moles halve as well, which means the enthalpy change of neutralization (i.e. Q) halves as well. As a result, ∆T goes down by 2. So the net change is no difference in ∆T

 

Basically ∆T = 2Q/(2mc), and the 2's cancel

 

Hope this helped