Question

The heat of combustion (Δh) for propane (molar mass 44.11 g/mol) is -2222 kj/mol. The heat generated from burning 1.680 g of propane is used to heat 575.0 g of water in a calorimeter. What is the change in temperature (°C) of the water?

Answer 1

To solve this problem, first calculate the moles of propane used. Next, calculate the heat energy released. Finally, use the formula q = mcΔT to calculate the change in temperature of the water.

Step-by-step explanation:

To solve this problem, we need to use the equation:

C3H8 + 5O2 → 3CO2 + 4H2O

First, calculate the moles of propane used:

1.680 g C3H8 * (1 mol C3H8 / 44.11 g) = 0.038 mol C3H8

Next, calculate the heat energy released:

0.038 mol C3H8 * -2222 kJ/mol = -84.4 kJ

Finally, use the formula q = mcΔT to calculate the change in temperature of the water:

-84.4 kJ = (575.0 g)(4.18 J/g°C)(ΔT)

ΔT = -84.4 kJ / (575.0 g * 4.18 J/g°C) = -0.0305°C

Since the change in temperature is negative, it means that the water will decrease in temperature by approximately 0.0305°C.