***When a 3.88 gram sample of solid ammonium nitrate dissolves in 60.0 grams of
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Question
***When a 3.88 gram sample of solid ammonium nitrate dissolves in 60.0 grams of water in a coffee-cup calorimeter, the temperature drops from 23.0°C to what temperature in °C. Specific heat of the solution = 4.18 J/g-K.
NH4NO3(s)-----> NH4 1+(aq) + NO3 1- (aq) H = + 25.3 kJ / mole NH4NO3
***A calorimeter contains 150.0 grams of water at 25.1°C. A 121.0 gram block of copper metal is heated to 100.4°C. The copper is added to the calorimeter. What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter. Specific heat of Cu(s) = 0.386 J/g-K. Specific heat of water = 4.18 J/g-K.
***A 2.200 gram sample of quinone, C6H4O2, is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/°C. The temperature of the calorimeter increases from 23.44°C to what temperature in °C. H of combustion = - 2745 kJ / mole quinone.
Explanation / Answer
1) q absorbed = m*s*DT
q = (3.88/80.0434)*25.3 = 1.23 kj
1.23*10^3 = 63.88*4.18*(x-23)
x = initial temperature = 27.6 C
2)
mcopper *s*DT = mwater*s*DT
(121*0.386*(100.4-x)) = (150*4.18*(x-25.1))
x = final temperature of mixer = 30.3 c
3)
Q released = 2745*(2.2/108.10) = 55.865 kj
55.865*10^3 = 7.854*10^3*(x-23.4)
x = final temperature = 30.51 C
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