The following information is given for benzene at 1 atm: boiling point = 80.10 °chvap(80.10 °c) = 393.3 j/gmelting point = 5.50 °chfus(5.50 °c) = 127.4 j/gspecific heat gas = 1.040 j/g°cspecific heat liquid = 1.740 j/g°ca 24.90 g sample of liquid benzene is initially at 45.70 °c. if the sample is heated at constant pressure (p = 1 atm), kj of energy are needed to raise the temperature of the sample to 103.00 °c.

11.87 kJ

Explanation:

First, the temperature must rise till the boiling point, with a gain of sensitive heat (without change of physical state). So the heat can be calculated by:

Q = mxClxΔT, where Q is the heat, m the mass, Cl the specific heat of the liquid, and ΔT the temperature variation:

Q1 = 24.90x1.740x(80.10 - 45.70)

Q1 = 1490.4 J = 1.49 kJ

Then, the liquid will be boiled, with a gain of latent heat, and the temperature must be constant. It will be:

Q = mxHv, where Hv is the heat of vaporization, so:

Q2 = 24.90x393.3

Q2 = 9793.17 J = 9.79 kJ

Then the temperature must increase to the final temperature, with a gain of sensitive heat:

Q3 = mxCgxΔT

Q3 = 24.9x1.040x(103.00 - 80.10)

Q3 = 593 J = 0.59 kJ

The total heat needes is: Q1 + Q2 + Q3 = 11.87 kJ

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