Question

A 5.000 g sample of methanol, C130H, was combusted in the presence of excess oxygen in a bomb calorimeter conaining 4000 g of water. The temperature of the water increased from 20.000 to 29.765 °C. The heat capacity of the calorimeter was 26571/°C. The specific heat of water is 4.184 Voc. Calculate for the reaction inkl/mol.

Answer 1

To calculate the heat of the reaction, we need to determine the heat absorbed by the water and the bomb in a bomb calorimeter. Once we have the values for qwater and qbomb, the heat of the reaction can be calculated using the equation ΔH = -qwater - qbomb.

Step-by-step explanation:

When a 5.000 g sample of methanol, CH3OH, is combusted in the presence of excess oxygen in a bomb calorimeter containing 4000 g of water, the temperature of the water increases from 20.000 to 29.765 °C. The heat capacity of the calorimeter is 26571/°C and the specific heat of water is 4.184 J/g °C. To calculate the heat of the reaction, we need to determine the heat absorbed by the water and the bomb.

1. Heat absorbed by water: qwater = mass of water × specific heat of water × temperature change
2. Heat absorbed by the bomb: qbomb = heat capacity of the bomb calorimeter × temperature change

Once we have the values for qwater and qbomb, the heat of the reaction can be calculated using the equation:

ΔH = -qwater - qbomb