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What amount of heat energy in Btus is required to raise the temperature of 45 pounds of lead from 62 degrees F to 100 degrees F?
Wiki User
∙ 15y ago
You like the old units! I'll work in SI then give you the answer in BTU. Specific heat capacity of lead = 0.127 KJ.Kg-1.C-1 (kilojoules per kilogram per deg C). 45 lb = 20.41 kg.
100 - 62 = 38 deg F = 21.11 degC.
So heat energy = 0.127 x 20.41 x 21.11 = 54.72 KJ. 1 KJ = 0.9478 BTU
Wiki User
∙ 15y ago
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