Question

If a 1.455 g sample of a new organic material is combusted in a bomb calorimeter and the temperature increases from 24.45°C to 29.74°C, what is the heat of combustion per gram of the material?

a) 2.37 kJ/g
b) 4.68 kJ/g
c) 31.61 kJ/g
d) 43.29 kJ/g

Answer 1

To determine the heat of combustion per gram, we need the calorimeter's heat capacity, which is not provided. Without this information, we cannot accurately perform the calculation to find the heat of combustion for this particular organic compound.

Step-by-step explanation:

To calculate the heat of combustion per gram of the material, we need to determine the total heat released during the combustion and then divide it by the mass of the sample. Assuming the bomb calorimeter and its contents have a known heat capacity, which is not stated in the question but is necessary for calculations, the heat absorbed (№) can be calculated using the formula № = heat capacity × ΔT, where ΔT is the change in temperature.

However, since the heat capacity is not provided in this example, we cannot proceed with an accurate calculation. In an actual problem like this, the heat capacity of the calorimeter must be known or provided to determine the total heat released.

If we had the heat capacity, we would follow these steps:

1. Calculate the total heat absorbed by the calorimeter using the formula № = heat capacity × (ΔT).
2. Divide the total heat absorbed by the mass of the organic sample to find the heat of combustion per gram.

Without the calorimeter's heat capacity, we can't Select an answer, but we would look for an answer roughly calculated using the difference in temperature multiplied by an assumed heat capacity and then divided by the mass of the sample (1.455 g).