Question #f0ab7

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Question #f0ab7

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Answer 1 · 2017-02-22T10:01:50

$3.96 \times 10^{5} J$ $3.96xx10^5J$ , rounded to two decimal places.

Explanation:

Taking Specific heat of Aluminum as $904 J {(k g K)}^{- 1}$ $904J(kg K)^-1$

Heat required is given by
$DeltaQ=mst$ ,
where $m,s and t$ are the mass, specific heat and rise or gain in temperature of the object; and for change of state as
$DeltaQ=mL$
where $L$ is the latent heat for the change of state and
Total heat to be added is $"Heat required to raise the temperature of solid aluminum from "130^@C " to " 600^@C+"Heat required to melt it"$
$=mst+mL$
$=0.45xx904xx(660-130)+0.45xx4.0xx10^5$
$=215604+1.8xx10^5$
$=3.96xx10^5J$ , rounded to two decimal places.

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