That 25.7 kJ usually is quoted as 235.7 kJ/mol which means 25.7 kJ/80g.
Therefore, heat absorbed by the NH4NO3 will be 25,700 J x (40/80) = about 13,000 J
Then -13000J = 125 x 4.184 x (Tf-Ti)
Tf = unknown
Ti = 25C
Tf is something like 0.5C but that's an estimate.
NH4NO3(s) + H2O(l) NH4NO3(aq) ΔH = +25.7 kJ
What is the final temperature in a squeezed cold pack that contains 40.0 g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18 J/(g·°C) for the solution, an initial temperature of 25.0°C, and no heat transfer between the cold pack and the environment.
For this question im suppose to use the q=mc delta T ?
1 answer
1 answer