Calorimetry+Problems

Where can we find the info for the pre-labs? Is it in the textbook?

Post your calorimetry problems' answers here!

1. __Kettle__ mk = 1kg =1000g ck = 0.385 kJ/kg deg. C = 0.385 J/g deg. C Tik = 20 deg. C (this is inferred) Tfk = 80 deg. C delta Tk = 60 deg. C (this is inferred)

__Water__ mw = 0.5kg = 500g cw = 4.18 J/g deg. C Tiw = 20 deg. C Tfw = 80 deg. C delta Tw = 80 deg. C

Qw = (mw) (cw) (delta Tw) Qw = (500g) (4.18 J/g deg. C) (60 deg. C) Qw = 125 400 J = 125.4 kJ

Qk = (mk) (ck) (delta Tk) Qk= (1000g) (0.385 J/g deg. C) (60 deg. C) Qk = 23100 J = 23.1 kJ

Qw + Qk = Qtotal 125.4 kJ + 23.1 kJ = 148.5 kJ

Therefore the amount of energy needed to raise the temperature of the kettle and its contents to 80 deg. C is 148.5 kJ.

2. Qneut = 57 kJ/mol (aka 57 kJ heat released per mole)

a) Required to find: Q released after burning 50mL, 1.0M HCl and 50mL, 1.0M NaOH First, write the chemical equation. Then write the information under the respective molecule so we know what is coming from each. HCl(l) + NaOH(s) à NaCl(s) + H2O(l) 50mL 50mL 1M 1M

Since the molar means “concentration”, we can multiply the molar by the volume (in L, don’t forget to convert from mL) to find the number of moles of the reactants:

C[M (or mol/L)] = moles/volume n = 1M / 0.05L n = 1 mol/L / 0.05L n = 0.05 mol

Heat is measured in J or kJ per mole. We know the kJ/mol from the question (57 kJ/mol) for one mole. We have 0.05 moles. So we divide 57kJ/mol by 0.05mol.

Q = 57kJ/mol / 0.05mol Q = 2.85 kJ

b) We are now required to find the final temperature given a starting temperature of 20°C. We can use the specific heat capacity and masses or water since these are in a solution. We know Q, m, c, and Ti, so just plug things in! The 100g came by adding 50mL and 50mL of reactants, and assuming density of water we can conclude that the mass is 100g. Don’t forget to convert the kJ to J as well.

Q = m c (Tf – Ti) 2850J = (100g) (4.18J/g°C) (Tf­­­­­­ - 20°C) 2850 / (100) (4.18)= (Tf­­­­­­ - 20) Tf­­­­­­ - 20 = 6.818 Tf­­­­­­ = 20 + 6.818 Tf­­­­­­ = 26.8°C

3. __Methane(system)__ NCH4=1 mol Qsys= -900KJ __ Water (surroundings) __ Qw= +900KJ mw= 100L = 100,000 mL =10,0000g Ti= 20˚C Tf =45˚C ΔT = (45-20) ˚C = 25˚C

Qw= mwcw ΔTw = 10,0000g x 4.18 J/g˚C x 25˚C = 10,450000 J = 10450 KJ methane would need to release -10450KJ of energy to heat up 100L of water Qsys = -10450 KJ 1 mol of CH4 / -900 KJ = ?mol CH4 / -10450 KJ No of mol of CH4 = 1mol CH4 x -10450 KJ / -900KJ No of mol of CH4 needed = 11.61111 mol Mass of CH4 = n x mm  = 11.6111 mol x 16.05 g/mol = 186.3 g of CH4 is needed Therefore we need 186.3g of CH4 to heat 100L of water.