If 3.65g of butane is burned underneath a cup holding 1.00 L of water at 21.0oC,what will the final temperature of the water be (ΔHcomb = -3325 kJ/mol)? (3 marks)

To solve this problem, we need to follow these steps:

Step 1: Convert the mass of butane to moles. The molar mass of butane (C4H10) is approximately 58.12 g/mol. So, 3.65g of butane is 3.65g / 58.12 g/mol = 0.0628 mol.

Step 2: Calculate the energy released by the combustion of butane. The heat of combustion (ΔHcomb) is the energy released when one mole of a substance is completely burned in oxygen. For butane, ΔHcomb = -3325 kJ/mol. So, the energy released by burning 0.0628 mol of butane is 0.0628 mol * -3325 kJ/mol = -208.7 kJ. The negative sign indicates that energy is released (exothermic reaction).

Step 3: Calculate the heat absorbed by the water. The specific heat capacity of water is 4.18 J/g°C. The mass of 1.00 L of water is approximately 1000 g. The heat absorbed by the water (q) can be calculated using the formula q = mcΔT, where m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature. We can rearrange this formula to solve for ΔT: ΔT = q / (mc).

Step 4: Convert the energy released by the combustion from kJ to J: -208.7 kJ = -208700 J.

Step 5: Substitute the known values into the formula for ΔT: ΔT = -208700 J / (1000 g * 4.18 J/g°C) = -49.9°C.

Step 6: Add the change in temperature to the initial temperature to find the final temperature: 21.0°C + (-49.9°C) = -28.9°C.

However, this answer doesn't make sense because the temperature of the water can't decrease when it's heated by burning butane. There must be a mistake in the calculation. Let's check the calculation again.

The mistake is in Step 5. The heat absorbed by the water should be positive because the water is heated by the burning butane. So, ΔT should be calculated as ΔT = 208700 J / (1000 g * 4.18 J/g°C) = 49.9°C.

So, the correct final temperature of the water is 21.0°C + 49.9°C = 70.9°C.