Calorimetry Problems

1.                  Phileas Fogg, the character who went around the world in 80 days, was very fussy about his bathwater temperature. It had to be exactly 38.0^o C. You are his butler, and one morning while checking his bath temperature, you notice that it’s 42.0^oC. You plan to cool the 100.0 kg of water to the desired temperature by adding an aluminum-duckie originally at freezer temperature (-24.0^oC). Of what mass should the Al-duckie be? [Specific heat of Al = 0.900 J/(g^oC); density of water =1 .00 g/ml]. Assume that no heat is lost to the air.

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.

2.                  A certain material’s(environment) temperature increases by 1.0^oC for every 1560 J that it gains. A 0.1964 g sample of quinone (molar mass = 108.1 g/mole) was burnt, and the surrounding material’s temperature increased from 20.3 ^oC to 23.5 ^oC. Find the molar heat of combustion for quinone.

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

3.                  A 1.55 g of CH₄O sample is burnt in a calorimeter. If the molar heat of combustion of CH₄O is -725 kJ/mole, and assuming that the 2.0 L of water absorbed all of the heat of combustion, what temperature change did the water experience?

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

4.                  0.20 moles of HX were neutralized by NaOH. The concentrations of the base and acid were equal. If the temperature of the water in the calorimeter increased from 19.9 to 24.6 C, what was the original concentration of HX?                              

            Molar heat of neutralization = -80 kJ/mole of HX

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.

5.                  In real calorimeters, most of the heat released by the bomb is absorbed by water, but a certain amount is also absorbed by the metal and insulation surrounding the water tank. A certain calorimeter absorbs 24 J/^oC. If 50.0 g of 52.7^oC water is mixed with the calorimeter’s original 50.0 g of 22.3^oC water, what will be the final temperature of the mixture?

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.