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What amount of heat energy is required to raise the temperature of 45 pounds of lead from 62 degrees to 100 degreesF?
Wiki User
∙ 13y ago
Specific heat of lead = 0.160 J/gC
45 pounds (454 grams/ 1 pound)
= 20430 grams
Use.
q(Joules) = mass * specific heat * change in temp.
q = (20430 grams)(0.160 J/gC)(100 C - 62 C)
= 1.2 X 10^5 Joules
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or
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1.2 X 10^5 Joules (1 calorie/4.184 Joules)
= 29688 calories
Wiki User
∙ 13y ago
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