Answer:
Cs * M of solution * deltaT = -(molesNH4NO3)*DeltaH of reaction
4.18 J/(g·°C) * (50.0 g + 115 mL * 1.0 g/mL) * deltaT = -(50.0 g/80.04 g/mol)*25.7 kJ
deltaT = -25.7 kJ / (4.18 J/(g·°C) * (50.0 g + 115 mL * 1.0 g/mL))
deltaT = -25.7 kJ / (4.18 J/(g·°C) * (165.0 g))
deltaT = -25.7 kJ / 690.7 J/°C
deltaT = -37.1°C
Final temperature = 25.0°C - 37.1°C = -12.1°C
Instant cold packs used to treat athletic injuries contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the following endothermic reaction.
NH4NO3(s) + H2O(l) = NH4NO3(aq) ÄH = +25.7 kJ
What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 115 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. To find the mass of water use the density of water = 1.0 g/mL. Hint: The process takes place at constant pressure.
So i know at constant pressure i have to use
Cs * M of solution * deltaT = -(molesNH4NO3)*DeltaH of reaction
Can someone please set this problem up because I tried to set it up but had difficulties with some of the values.
1 answer
1 answer