Answer on Molecular Physics | Thermodynamics Question for Bila hussa
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.
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