A 0.468 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water in it rose from 20.45 °C to 23.65 °C. The heat capacity of the calorimeter by itself is 2.21 kJ/°C and the specific heat capacity of water is 4.184 J/g.°C What is the heat of combustion per mole of pentane?

E = 3,154.37 KJ

The heat of combustion per mole of pentane is 3,154.37 KJ

Explanation:

Given;

Change in temperature of system ∆T = 23.65-20.45 = 3.2 °C

Mass of water m1 = 1 kg

Specific heat capacity of water C1 = 4.184J/g°C = 4184J/kg °C

Heat capacity of calorimeter mC2 = 2.21 kJ/°C = 2210J/°C

Heat gained by both calorimeter and water is;

H = (m1C1 + mC2)∆T

Substituting the values;

H = (1×4184 + 2210)×3.2

H = 20460.8 J

Mass of pentane burned = 0.468 g

Molecular mass of pentane = 72.15g

If 0.468g of pentane releases 20460.8 J of heat,

72.15g of pentane will release;

E = (72.15/0.468) × 20460.8 J

E = 3154373.333333J

E = 3,154.37 KJ

The heat of combustion per mole of pentane is 3,154.37 KJ

electric field is a function of 1/r^2

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