What would be the temperature fall?

When 0.10 mol of ammonium nitrate is dissolved in 100 cm3 of water, the temperature falls by 5.0 K. What would be the temperature fall when 0.02 mol of ammonium nitrate is dissolved in 10 cm3 of water, under the same conditions?

A 1.0 K
B 2.0 K
C 5.0 K
D 10.0 K

The right answer is D but I have no idea how to solve this...

Original post by ja.zii

When 0.10 mol of ammonium nitrate is dissolved in 100 cm3 of water, the temperature falls by 5.0 K. What would be the temperature fall when 0.02 mol of ammonium nitrate is dissolved in 10 cm3 of water, under the same conditions?

A 1.0 K
B 2.0 K
C 5.0 K
D 10.0 K

The right answer is D but I have no idea how to solve this...

q = mcΔT

so

ΔT= q/mc

ΔTis proportional to q/m

c is a constant, so can be ignored when dealing with proportional change.

In this case there is 1/5 the number of moles so q would be 1/5 of the original value BUT
the mass is 1/10 the mass in the second case and so you would expect the heat change to be 10x greater.

10 x 1/5 = 2

so you would expect the temperature change to be double.