Question

access important values if needed for this question.

A piece of solid lead weighing 38.4 g at a temperature of 315 °C is placed in 384 g of liquid lead at a temperature of 372 °C. After a while, the solid melts and a completely liquid sample remains. Calculate the temperature after thermal equilibrium is reached, assuming no heat loss to the surroundings.

The enthalpy of fusion of solid lead is ΔH_fus = 4.77 kJ/mol at its melting point of 328 °C, and the molar heat capacities for solid and liquid lead are C_solid = 26.9 J/mol K and C_liquid = 28.6J/mol K.

T_final =  °C

Answer 1

The amount of energy required q1 to heat 38.4 g of lead at 315 C to 328 C = Csolid*moles*T

Where T = final temperature - initial temperature

= 328 C - 315 C = 13 C

moles of lead in 38.4 g = weight/molar mass

= 38.4g / 207.2 g/mole = 0.185 moles

Csolid = 26.9 J/mole

so, q1 = 26.9 Jmole*0.185 mole * 13 C = 64.7 J

Now, the melting of this solid will occur so, amount of heat q2 required to melt 0.185 moles of lead

q2 = Hvap*mole = 4.77 kJ/mole*0.185 moles

= 0.88245 kJ = 0.88245 * 1000 J = 882.45 J

Now, lets say after melting the final temperature of the melted lead is x C so,

heat required to attain temperature of x C after melting at 328 C is q3

q3 = Cliquid*moles*(x-328)

q3 = 28.6 J/molK * 0.185 moles * (x C - 328 C)

q3 = 5.291*(x-328) J

These energies q1,q2,q3 must be provided to do this conversion.

This energy must be taken from the liquid lead and thus its temperature will go down.

So, the heat required Q = q1 + q2 + q3

and Q = Cliquid*moles*(372-x)

moles in 384g of lead = weight/molar mass

= 384g/207.2 g/mole = 1.85 moles

so, Q = 28.6 J/mol-K * 1.85 moles * (372 C-x C)

Q = q1 + q2 + q3

28.6*1.85*(372-x) = 64.7 J + 882.45 J + 5.291*(x-328) J

19682.52 - 52.91x = 947.15 +5.291x - 1735.448

58.2x = 20470.818

so, x = 20470.818/58.2 C = 351.73 C

So, final temperature will be 351.73 C