Check your work! (Solutions to Calorimetry Problems)

Solution #1

-Q_water = Q_Al

m_wc_wDT_w = - m_Alc_AlDT_Al

100 000g(4.19 J/[g C] (38.0 – 42.0)= - m_Al (0.900 J/[g C])(38.0 – [-24.0]).

m_Al = 30036 g = 30.0 kg.

Solution #2

Q_material = 1560 J/^oC (23.5-20.3) ^oC  = 49 92 J = 5.0 kJ

Note that 1560 J/^oC is equivalent to mc from Q = mcDT.

DH = -Q = -5.0 kJ

n = 0.1964g/[108.1 g/mole] = 0.00182 mole

DH/n = -5.0 kJ/0.00182 mole = -2.7 X 10³ kJ/mole of quinine

Solution #3

n = 1.55g/[32g/mole] = 0.0484 moles

n[DH/n] = DH

0.0484 moles(-725 kJ/mole)= - 35.1 kJ

Q = - DH

Q = 35.1 kJ = 35 100 J

Q = mc DT

35 100 = 2000g (4.19 J/[g C]) DT

DT = 4.2 C

Solution #4

n[DH/n] = DH

0.20mole [-80 kJ/mole] = -16 kJ.

Q = - DH

Q = 16 kJ = 16000 J.

Q = mc DT

16000 = m (4.19 J/[g C])( 24.6 – 19.9)

m = 812g of water.

Since the concentrations of the acid and the number of moles were equal (moles are equal because of the 1:1 ratio in which HX reacts with NaOH), then the volumes used were equal.

812 g = 812 mL created from 812/2 = 411 mL of acid = 0.411 L

concentration = n/V = 0.20 moles/0.411 L = 0.49 M.

Solution #5

The heat lost by the hot water will be gained by the cold water and by the calorimeter.

-Q_hot = Q_cold + Q_calor

-mc DT = mc DT + 24J/⁰C DT

-50(4.19)( x – 52.7) = 50(4.19)(x –22.3) + 24(x – 22.3)

-209.5( x – 52.7) = 233.5(x –22.3)

-0.897( x – 52.7) = x – 22.3

-0.897 x  + 47.27 = x – 22.3

69.57 = 1.897 x

       x = 37 ^oC.