Question

What final temperature would you expect from a chemical ice pack with an initial temperature of 25.0 ^oC that mixes 30.00 g of KNO₃ (∆H = 34.89 kJ/mol) with 200.0 g water? Assume that the specific heat of the solution is 4.18 J/g^oC.

Answer 1

moles of KNO3 = mass / molar mass

                         = 30.0 / 101.1

                         = 0.297

∆H   = - Q / n

34.89 = - Q / 0.297

Q = -10.353 kJ = -10353 J

Q = m Cp dT

-10353 J = 230 x 4.18 x dT

dT = -10.77

T2 - T1 = -10.77

T2 - 25 = -10.77

T2 = 14.2 oC

final temperature = 14.2 ^oC