Answer to Question #174375 in Chemistry for Ava Brooke

Question

A 27.98 g sample of aluminum is heated from 25 ^oC to its melting point of 660 ^oC and then melted completely. How much energy was absorbed by the sample? (Answer in kJ)

S=0.91 J/g^oC Delta H_f = 10.7 kJ/mol

Answer 1

Solution:

The molar mass of Al is 26.981 g/mol.

Hence,

(27.98 g Al) × (1 mol Al / 26.981 g Al) = 1.037 mol Al

Moles of Al = 1.037 mol

This problem can be summarized thusly:

1) A sample of aluminum is heated from 25°C to 660°C. The heat absorbed is calculated by using the specific heat of aluminum (S = 0.91 J/g°C) and the equation: q = m × S × ΔT.

q₁ = m × S × (T_f - T_i)

q₁ = 27.98 g × 0.91 J/g°C × (660 - 25)°C = 16168.243 J = 16.168 kJ

q₁ = 16.168 kJ

2) A sample of aluminum is melted at 660°C. The heat absorbed is calculated by multiplying the moles of aluminum by the molar heat of fusion (ΔH_f).

q₂ = ΔH_f × n = 10.7 kJ/mol × 1.037 mol = 11.096 kJ

q₂ = 11.096 kJ

q = q₁ + q₂ = 16.168 kJ + 11.096 kJ = 27.264 kJ = 27.26 kJ

q = 27.26 kJ

Answer: 27.26 kJ of energy was absorbed by the sample.