Question

4. 100.0 mL of water is placed in a calorimeter and allowed to reach an initial temperature of 23°C. A 45.0g block of Cu (SH=0.385 J/g at 285°C is placed in the water (SH = 4.18 J/g. C) in the calorimeter. a) If the final temperature reached is 30°C, determine the heat capacity of the calorimeter? b) The same calorimeter was used to determine the heat of solution of 20.0g of NH4NO3 NH4NO3 (s) + NH4NO3 (29) The initial temperature of the 100.0 mL of water and calorimeter were again 23°C. Once all of the NH4NO3 dissolved, the temperature reached was 11°C. Determine AH”solution for NH4NO3. Ansal/mol

Answer 1

4)

a)

Heat loss by copper = heat by water and calorimeter.

or

m1 * s1 * dT1 = m2 * s2 * dT2 * S3 (calorimeter) * dT2

or

45.0 * 0.385 * (285 - 30) = 100.0 * 4.18 * (30 - 23) + S3 (calorimeter) *(30 - 23)

or

4417.9 = 2926 + 7 * S3

or

S3 = 213.1 J / oC

thus

heat capacity of calorimeter = 213.1 J / oC

b)

mole of NH4NO3 = 20.0 g / 80.043 g/mole = 0.2499 mole.

Heat = m2 * s2 * dT2 * S3 (calorimeter) * dT2

or

Heat = 100.0 * 4.18 * (11 - 23) + 213.1 *(11 - 23) = - 7573.2 J

thus

delta H (solution) = 7573.2 J / 0.2499 mole = 30305 J / mole = 30.3 KJ / mole.