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Calculate the number of k.J of heat necessary to raise the temperature of \(60.0 \mathrm{~g}\) of aluminium from \(35^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\). Molar heat capacity of \(\mathrm{Al}\) is \(24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\).
Solution:
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Verified Answer
\(\mathrm{q}=\mathrm{n} \times \mathrm{C} \times \Delta \mathrm{T}\)
\(\begin{aligned}
&=\left(\frac{60 \mathrm{~mol}}{27}\right) \times\left(24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right) \times(328-308) \mathrm{K} \\
&=1066.7 \mathrm{~J}=1.07 \mathrm{~kJ}
\end{aligned}\)
\(\begin{aligned}
&=\left(\frac{60 \mathrm{~mol}}{27}\right) \times\left(24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right) \times(328-308) \mathrm{K} \\
&=1066.7 \mathrm{~J}=1.07 \mathrm{~kJ}
\end{aligned}\)
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