what is the amount of heat energy required to raise 1kg of lead to 10degree celcius
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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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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
Hope this information will clear your doubt about heat energy.
If you have any doubt do ask on the forum our experts will help you out.
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