Answer:
Information shadow:
Heat capacity 'C of
\[Al=24Jmo{{l}^{-1}}{{K}^{-1}}\]
Temperature
rise, \[\Delta T=20\]
Mass of
aluminium = 60 g
Number of
moles of \[Al=\frac{60}{27}=2.22\]
Problem
solving strategy:
Heat
absorbed to raise the temperature of n mole \[\text{ }\!\!'\!\!\text{ Al
}\!\!'\!\!\text{ }\] by \[\Delta T\]may be calculated as,
\[q=nC\Delta T\]
Working
it out:
\[q=2.22\times
24\times 20=1065.6J\]
= 1.065 kJ
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