THERMO Spoken Here! ~ J. Pohl ©  ( D6800~2/15)( D7280 -  Specific Heat Calculation)

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.

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. What least heat is required to melt all of the lead?

Premise presently unwritted!