(1) Enthalpy of fusion is given as \(2.04*10^4\frac{J}{kg}\) . It means \(2.04*10^4\) J of heat is required to melt 1 kg of lead. We are asked to calculate the heat required to melt 2.5 kg of lead.
\(q=m\Delta H_f\)
where m is the mass and \(\Delta H_f\) is the enthalpy of fusion.
let's plug in the values in the formula:
\(q=2.5kg*2.04*10^4\frac{J}{kg}\)
\(q=5.1*10^4J\)
(2) From given heat energy and mass we can calculate the enthalpy of fusion and then match it with the values given in the table and see which metal is it.
\(\Delta H_f=\frac{q}{m}\)
\(\Delta H_f=\frac{2.52*10^5J}{4.00kg}\)
\(\Delta H_f=6.3*10^4\frac{J}{kg}\)
Looking at the standard table, the substance is Gold as it has same enthalpy of fusion value that we have calculated.
(3) Heat of vaporization is used when the liquid changes into vapor. Heat of condensation is used when the vapor changes into liquid and the heat of solidification is used when the liquid freezes.
In question it asks, what would be used to calculate the heat released when a liquid freezes. So, the right option is "Heat of solidification."
(4) This is exactly Similar to the first one, the only difference is, here we will use the enthalpy of vaporization.
\(q=m\Delta H_v\)
\(q=0.200kg*2.26*10^3\frac{kJ}{kg}\)
\(q=4.52*10^2kJ\)
First option is correct.
