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Question
650 J of heat is required to raise the temp. of 0.25 kg of lead from 15°C to 35°C. Calculate the Sp. heat capacity of lead.
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Solution
Q = 650 J
m = 0.25 kg
ΔT = (35 - 15) = 20°C
Q = m x C x T
C = `"Q"/("m" xx Δ"T") = 650/(0.25 xx 20) = 130` J/kg°C
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