- #1
ally1h
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Homework Statement
A sculptor has come to you with a problem, because she knows that you are such a good physics student. She wishes to make a lead casting of an object which will require 2 kg of melted lead. If the lead was stored in her basement at approximately 55˚F, and it is to be melted in a 1/4" thick steel ladle/pot of approximate cross sectional area 12in^2, how long will it take to melt the lead if it is placed in a kiln at 617.5˚F, the melting point of lead?
Lf = 0.25x10^5 J/kg
Lv = 8.7x10^5 J/kg
c = 130 J/kg*C˚
k = 40 J/s*m*C˚
Homework Equations
P = kA(ΔT/d)
Q = mcΔT
Q = mL
P = Q/Δt
The Attempt at a Solution
12in^2 = 0.0077 m^2
55˚F = 12.7˚C
617.5˚F = 325.3˚C
0.25 in = 0.00635 m
P = kA(ΔT/d)
P = (40 J/s*m*C˚)(0.0077m^2)((325.3˚C-12.7˚C) / 0.00635 m)
P = 15162.3
Q1 = mcΔT
Q1 = (2 kg)(130 J/kg*˚C)(325.3˚C-12.7˚C)
Q1 = 81276 J
Q2 = mL
Q2 = (2 kg)(0.25x10^5 J/kg)
Q2 = 50000 J
Q1+Q2 = 81276 J + 50000 J = 131,276 J
P = Q/Δt
15162.3 = (131,276 J)t
t = 0.1155 s
So... I REALLY don't think this is correct, but I can't seem to think of another way around this. Any help would be appreciated!