Answer to Question #112369 in General Chemistry for Budamala

Question

A 53.5 gram sample of solid mercury at -38.83 degrees celcius is heated to gaseous mercury at 356.73 degrees celcius. How much energy was added? melting point= -38.83 degrees celcius; boiling point= 356.73 degrees celcius

Answer 1

Solution.

m = 53.5 g = 53.5x10^-3kg;

t₁ = -38.83^oC;

t₂ = 356.73^oC;

t_melting=-38.83^oC;

t_boiling = 336.73^oC;

c = 140J/(kg $\sdot$^oC);

$\lambda$ = 12 x 10³ J/kg;

L = 0.3 x 10⁶ J/kg;

Q = Q₁ + Q₂ + Q₃ ;

Q₁ = $\lambda$ m - mercury melting;

Q₂ = cm$\Delta$t - heating the liquid mercury to the evaporation temperature;

Q₃ = Lm - evaporation of mercury;

Q₁ = 12 x 10³ J/kg x 53.5x10^-3kg = 642 J;

Q₂ = 140 J/(kg ⋅^oC) x 53.5x10^-3kg x ( 356.73^oC - (-38.83^oC)) = 2962.74 J;

Q₃ = 0.3 x 10⁶ J/kg x 53.5x10^-3kg = 16050J;

Q = 642 J + 2962.74 J + 16050J = 19654.74 J = 19.65474 kJ;

Answer: Q = 19.65474 kJ.