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What is the amount of heat energy required to raise 1kg of lead to 10degree celcius

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Kindly find the ANSWER of your asked query.
Heat energy = ML+(MCV(∆T))water+(mCv(∆T))iceQ= 777000 JL = Q−(mCv(∆T))water+(mCv(∆T))icemL = 777000−1×4200×(100−(−10))−1×2100×1101=777000−462000−231000=84000 Joule


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