Question

Given the information shown in process "B", calculate the molar enthalphy of fusion for ice

Answer 1

Answer:

D) 6.02 kJ/mol

Explanation:

Hello,

In this case, since 200kJ melted 1.5 kg of ice (2.0kg-0.5kg), we can compute the melted moles:

$n=\frac{1.5kg}{18kg/kmol} *\frac{1000mol}{1kmol}=83.33mol$

Then, we compute the molar enthalpy of fusion by diving the melted moles to the applied heat:

$\Delta H_{fusion}=\frac{500kJ}{83.33mol}\\ \\\Delta H_{fusion}=6.02kJ/mol$

Hence answer is D) 6.02 kJ/mol.

Best regards.