Answer to Question #1646 in Molecular Physics | Thermodynamics for Bila hussa
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?
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 H1 = Cm m (T2-T1) = 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: H2 = 59.11kJ/mol * 2.49 mol = 147.34 kJ. The total heat is H1+H2 = 244.11 + 147.34 = 391.45 kJ.