Thermochemistry Summary

Question

Assuming the specific heat of the water to be 4.200 J g^-1 K^-1, calculate the change in temperature of 2.000 kg of water (assuming that only 63.15% of the energy is transferred to the water) that is heated by burning the butane in part (b).

Answer 1

Step 1: determine the heat liberated from burning C4H10 in part (b).

The combustion of C₄H₁₀ yield 288 kJ per mole. Calculate how many moles are in 4.00 g.

4.00 g C₄H₁₀ x 1 mol / 58 g = 0.0690 mols

Change in enthalpy = ∆H = 0.0690 mols x -288 kJ / mol = -19.9 kJ

Since only 63.15% of this is available to heat the water, that would be q = 19.9 kJ x 63.15% = 12.6 kJ

q = mC∆T

q = heat = 12.6 kJ

m = mass of water = 2.000 kg

C = specific heat of water = 4.200 kJ/kgº

∆T = change in temperature = ?

solving for ∆T...

∆T = q / mC

∆T = (12.6 kJ) / (2.000 kg)(4.200 kJ/kgº)

∆T = 1.5º