Question #1646

A 0.5 Kg sample of mercury at 240 K is to be brought to its boiling point, 337 C, and completely evaporated. how much heat would this take?

Expert's answer

Firstly we should convert the initial temperature to Celcium: 240 K = -33 C.

The specific heat of mercury is 1.25 kJ/kg·°C, therefore

H_{1} = C_{m} m (T_{2}-T_{1}) = 1.25 [kJ/kgC]*0.5 [kg]*(337+ 33)C = 244.11 kJ

In 0.5 kg there is n = 500[g]/200.59[g/mol] = 2.49 mol of mercury.

The heat required to vaporize of 2.49 mol of mercury is:

H_{2} = 59.11kJ/mol * 2.49 mol = 147.34 kJ.

The total heat is H_{1}+H_{2} = 244.11 + 147.34 = 391.45 kJ.

The specific heat of mercury is 1.25 kJ/kg·°C, therefore

H

In 0.5 kg there is n = 500[g]/200.59[g/mol] = 2.49 mol of mercury.

The heat required to vaporize of 2.49 mol of mercury is:

H

The total heat is H

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