Done 3) Near room temperature, the specific heat capacity of benzene is 1.15 a)

Question

Done 3) Near room temperature, the specific heat capacity of benzene is 1.15 a) Calculate the heat required to raise the temperature of 1545 mL of benzene from 9 °C to 55 °C. The density of liquid benzene is 0.78 mL b) 25.0 g of benzene is placed in a bomb calorimeter and pressurized with excess oxygen gas. The bomb calorimeter is placed in an insulated container containing 1000 g of water at 298 K and the benzene is combusted. If the molar heat of combustion of benzene is 399.7 kJ/mol, and assuming the calorimeter absorbs negligible heat, determine the final temperature of the water after the system has reached thermal equilibrium

Explanation / Answer

3.

a)

mass of benzene = density * volume = 0.78 * 1545 = 1205.1 gm.

Heat required = m * s * dT = 1205.1 * 1.15 * (55 - 9) = 63749.8 J = 63.75 KJ

so heat required is 63.75 KJ.

b)

20.0 gm benzene = mass / molar mass = 20.0 / 78 = 0.256 mole.

Heat released due to combustion of benzene = 0.256 * 399.7 = 102.487 KJ = 102487 KJ

Heat = m * s * (T2 - T1)

102487 = 1000 * 4.184 * (T2 - 298)

T2 = 322.5 K

so final temperature is 322.5 K

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