THERMO Spoken Here! ~ J. Pohl © ~ 2017 (D6800-211)

Leaded Pipe Joint

The joint or connection of two sections of cast iron pipe is accomplished by pouring and caulking lead. Preparing to "lead" a pipe joint, a plumber puts 15 pounds of lead into a "hot pot." The lead is initially 25°C Lead melts at 327°C. (clead = 0.13 J/g°C and hsf,lead = 23 J/g).

What least heat is required to melt all of the lead?
♦  This solution is reasonably direct:

ΔH = Q1-2

m(h2 - h1) = Q1-2

h2 = h1 + clead(327 - 25)°C + hsf

h1 = h1

h2 - h1 = clead(327 - 25)°C + hsf

15 lbm (kg/2.2 lbm){ 0.13 (kJ/kg°C)[327 -25]°C + 23 (kJ/kg)} = Q1-2

Q1-2 = 6.81 kg (39.3 + 23)(kJ/kg) = 424 kJ

Our solution tells us the "least" heat required. Were the event attempted in the Arctic with a strong wind blowing, the heat required would be much greater than our number.

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