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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!