Complete question:
A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.
Answer:
the heat of combustion of the methanol is 402.31 kJ/mol
Explanation:
Given;
mass of water, [tex]m_w[/tex] = 100 g
initial temperature of water, t₁ = 28 ⁰C
final temperature of water, t₂ = 58 ⁰C
specific heat capacity of water = 4.184 J/g⁰C
reacting mass of the methanol, m = 1.00 g
molecular mass of methanol = 32.04 g/mol
number of moles = 1 / 32.04
= 0.0312 mol
Apply the principle of conservation of energy;
[tex]n\Delta H_{methanol} = Q_{water}\\\\n\Delta H_{methanol} = mc\Delta t\\\\n\Delta H_{methanol} = 100 \times 4.184\times (58-28)\\\\n\Delta H_{methanol} = 12,552 \ J\\\\n\Delta H_{methanol} = 12.552 \ kJ\\\\\Delta H_{methanol} = \frac{12.552}{n} \\\\H_{methanol} = \frac{12.552 \ kJ}{0.0312 \ mol} \\\\\Delta H_{methanol} = 402.31 \ kJ/mol[/tex]
Therefore, the heat of combustion of the methanol is 402.31 kJ/mol
