Question

NH4NO3, whose heat of solution is 25.7 kJ/mol, is the one substance that can be used in a cold pack. If the goal is to decrease the temperature from 25 to 50 degrees Celsius, how many grams of NH4NO3 should we use for every 100.0g of water in the cold pack? Assume no heat was lost outside of the cold pack, and the specific heat of the resulted solution was the same as water, or 4.184 J/g degrees C.

Please include the steps.

Answer 1

m = 100.0 g

dT = 50 -25 = 25 oC

Cp = 4.184 J / oc g

Q = m Cp dT

Q = 100 x 4.184 x 25

Q = 10460 J

Q = 10.46 kJ

delta H = Q / n

25.7 = 10.46 / n

n = 10.46 /25.7

n = 0.407

moles of NH4NO3 = 0.407

mass of NH4NO3 = moles x molar mass

                               = 0.407 x 80

                               = 32.56 g

mass of NH4NO3 needed = 32.56 g