10 Points. Enthalpy change problem.

A coffee-cup calorimeter is calibrated by adding 1840 J of heat to the water in the calorimeter and measuring a 1.72 oC rise in temperature. When some NH4NO3(s) is added to the same water in the calorimeter, the temperature falls by 1.04 oC. Calculate the enthalpy change due to the dissolving of this NH4NO3(s).

I know the answer is +1110 J.

I need to know how to get it. This is the last one. I know it is probably simple, however, I am about burned out on this worksheet. Any help is greatly appreciated.

First part they are doing to find out the Heat capacity of the calorimeter
suppose it as S

you can use the equation Q = S(theta) where Q is the heat energy, S is the heat capacity and theta is the temperature change

from the first part you can find the S = Q / theta
= 1840 J / 1.72 Celsius = 1069.8

the u can use the same equation where you are going to calculate Q which is the enthalpy change
I hope you know the definition of enthalpy change ( the heat change under constant pressure)

so Q = S(theta)
= 1069.8 * 1.04
= 1112.59

I hope that is the answer you got

Since adding 1840J gives a 1.72 degree increase in temperature, a 1.04 degree increase would be calculated by 1.04/1.72 x 1840J = +1110J (when rounded to 3 s.f.)

Is this a job interview question so you can become a hipster barista at a indie coffee shop?

1.72 oC change = 1840 J
1.04 oC change = 1804/1.72 X1.04 = +1113J