I need both 3 and 4 please The enthalpy of combustion of benzoic acid (C_6H_5COO

Question

I need both 3 and 4 please The enthalpy of combustion of benzoic acid (C_6H_5COOH) is -3226 7 kJ/mol When 1 g of benzoic acid was oxidized in some constant-volume bomb calorimeter the temperature rose from 20.00 degree C to 22 00 degree C What would be the change in temperature if 1 g of glucose (C_6H_12O_6) is oxidized in the same calorimeter (To find the enthalpy of combustion of glucose, use standard molar enthalpies of formation from table 4 3 on page 103) The internal engine of n 1500 kg car is designed to run on octane (C_8 H_18), whose enthalpy of combustion is 5510 kJ/mol. Estimate the amount of fuel needed to climb a hill of 1 km height. The maximum temperature in the engine cylinder is 2000 degree C. the exit gases temperature is 700 degree C. Neglect all forms of friction.

Explanation / Answer

3) Enthalpy of combustion of benzoic acid = -3226.7 KJ / mole

so enthalpy of 122 grams = -3226.7 KJ

so energy of 1 gram = -3226.7 / 122 = 26.44 KJ

The raise in temperature = 2 0C

Q = Heat capacity X raise in temperature

So heat capacity = 26.44 KJ / 2 = 13.22 KJ

The Standard enthalpy of combusiton of glucose

-2808 kJ/mol

mol wt of glucose = 180

so energy liberated by combustion of 1 g of glucose = 2808 / 180 = 15.6 KJ

Again

Heat = heat capacity X rise in temperature

15.6 = 13.22 X rise in temperature

Rise in temperature = 1.18 0C

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