Answer to Question #174398 in Chemistry for Ava Brooke

Question

A 21.34 g sample of aluminum is heated from 28 ^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:

Moles of Al = Mass of Al / Molar mass of Al

The molar mass of Al is 26.981 g/mol.

Hence,

Moles of Al = 21.34 g / 26.981 g mol^-1 = 0.7909 mol = 0.791 mol

Moles of Al = 0.791 mol

This problem can be summarized thusly:

1) A sample of aluminum is heated from 28°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 = 21.34 g × 0.91 J/g°C × (660 - 28)°C = 12273.06 J = 12.273 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 × 0.791 mol = 8.464 kJ

q = q₁ + q₂ = 12.273 kJ + 8.464 kJ = 20.737 kJ = 20.74 kJ

q = 20.74 kJ

Answer: 20.74 kJ of energy was absorbed by the sample.