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If 42.0 kJ of heat is added to a 32.0g sample of liquid methant under 1 atm of pressure at atemperature of -170C, what are the final state and temperatrue of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is -161.5C. The specific heats of liquid and gaseous methane are 3.48 and 2.22 J/g*K, respectively. Heat of vaporization is 8.20 kJ/mol.

Respuesta :

meerkat18
The total heat considering both the sensible and latent heat is expressed by ΔH = (m Cp ΔT)liquid + (m ΔHv) + (m Cp ΔT)gas
42000 J = (32 g * 3.48 J/gK *(-161.5+170)K) + (32 g *(mol/16g) ( 8200 J/mol) + (32 g*2.22 J/gK (Tf+161.5) 
Tf = 185.54 C.

This is above the -161.5C boiling point, hence the final state is gas.

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