Answer:
The energy needed to melt 0.225 kg of lead with an initial temperature of 27.3 °C is 15.1563 kJ
Explanation:
To solve the question, we are required to use the properties of the lead, and the in this case the necessary properties are the heat capacity to rise the temperature of the lead to the melting point temperature, then at the melting point temperature, the latent heat required to melt the lead
The melting point of lead T₂ is 327.5 °C
The latent heat of fusion of lead, L = 4.799 kJ/mol
The specific heat capacity of lead, c = 0.13J/g K
The molar mass of lead, M = 207.2 g
Initial temperature T₁ = 27.3 °C
Mass of lead = 0.255 kg
Change in temperature from initial temperature, to melting point temperature = (T₂ - T₁) = ΔT
Heat required to melt the lead = H
Note that the latent heat is given in kJ/mol therefore in the following equation in the calculation for latent heat component we have m/M
Therefore to melt the lead we have H = Sensible heat component (m×c×ΔT) + latent heat component (m/M×L)
Therefore to melt the lead we have H = m×c×ΔT + m/M×L
= 0.255 kg × (1000 g/kg) × (327.5 - 27.5)×0.13J/g K + ((0.225 kg × 1000 g/kg)/(207.2 g))× 4.799 kJ/mol × 1000J/ kJ
9945.0 J +5211.3 J = 15156.3 J
The heat energy required to melt 0.225 kg of lead that was initially at 27.3 degrees Celsius is 15156.3 J or 15.1563 kJ
