Answer to Question #176212 in Chemistry for marina

Question

A 37.98 g solid sample of mercury is heated from -39 ^oC its melting point, completely melted then heated to 106 ^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 × 37.98 g = 0.418 kJ

q₁ = 0.418 kJ

2) A sample of liquid mercury is heated from -39°C to 106°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 = 37.98 g × 0.14 J/g°C × (106 - (-39))°C = 770.99 J = 771 J = 0.771 kJ

q₂ = 0.771 kJ

q = q₁ + q₂ = 0.418 kJ + 0.771 kJ = 1.189 kJ = 1.19 kJ

q = 1.19 kJ

Answer: 1.19 kJ of energy was absorbed by the sample.