# 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

• Your other proposition is correct. We have twice the amount of reagents, hence twice the heat, but also twice the volume of water. – Ivan Neretin Mar 13 '16 at 17:38
• @IvanNeretin, but why is my other proposition incorrect? – CCC Mar 13 '16 at 18:47
• Same amount of reagents means same amount of heat. But heat and temperature are not quite the same. Would that amount of heat warm any amount of water by $6^\circ$C? Would it, say, warm an ocean by $6^\circ$C? – Ivan Neretin Mar 13 '16 at 19:02

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