Answer to Question #174399 in Chemistry for Ava Brooke

Question

A 38.23 g solid sample of mercury is heated from -39 ^oC its melting point, completely melted then heated to 109 ^oC, How much energy was absorbed by the sample? (Answer in kJ)

S=0.14 J/g^oC Delta H_f = 0.011 kJ/g

Answer 1

Solution:

This problem can be summarized thusly:

1) A sample of solid mercury is melted at -39°C. The heat absorbed is calculated by multiplying the grams of solid mercury by the heat of fusion (ΔH_f) in kJ/g.

q₁ = ΔH_f × m = 0.011 kJ/g × 38.23 g = 0.4205 kJ

2) A sample of liquid mercury is heated from -39°C to 109°C. The heat absorbed is calculated by using the specific heat of liquid mercury (S = 0.14 J/g°C) and the equation: q = m × S × ΔT.

q₂ = m × S × (T_f - T_i)

q₂ = 38.23 g × 0.14 J/g°C × (109 - (-39))°C = 792.1256 J = 0.7921 kJ

q = q₁ + q₂ = 0.4205 kJ + 0.7921 kJ = 1.2126 kJ = 1.213 kJ

q = 1.213 kJ

Answer: 1.213 kJ of energy was absorbed by the sample.