The temperature rise after neutralisation The answer to this question is B, however, I don't quite understand.
If considering Q= mc delta T, the heat released is to some extent being influenced by the no. of moles. If using that line of thought, wouldn't we want C where it has also the same no. of moles?
If we really have to arrive at B, my other proposition would be that the increase in heat is dissipated in a larger volume, hence giving a temperature rise of 6 degrees C.
I am a bit confused by this question. Any comments would be very helpful. Thank you in advance
One calorie is the amount of energy (heat) needed to raise 1 gram of water 1 degree Celsius at a pressure of one atmosphere. (It is almost constant to 3 significant figures, but there is some temperature dependence...) Also assume that 1 cm^3 of water is 1 gram which is more than good enough for the calculations needed for the problem. 

So for the reaction in the problem given, 50 mL of water by 6 degrees is 300 calories.

For the other reactions I'll give calories and temperature rise for product solution. 

(a) 600 calories, 12 degrees C

(b) 600 calories, 6 degrees C

(c) 300 calories, 3 degrees C

(d) 300 calories, 1.5 degrees C