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The heat required to rise the temperature of $54 \mathrm{~g}$ of aluminium from $40^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in $\mathrm{J}$ is (molar heat capacity of aluminium in this temperature range is $24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$; atomic weight of $\mathrm{Al}$ is 27)
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Verified Answer
The correct answer is:
960
Given,
Weight of aluminium $(w)=54 \mathrm{~g}$
Temperature difference $(\Delta T)=60^{\circ}-40^{\circ}=20^{\circ} \mathrm{C}$
Molar heat capacity of aluminium
$$
\left(C_s\right)=24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
$$
Atomic mass of $(\mathrm{Al})$, i.e. $(M)=27$
$\because \operatorname{Moles}$ of $(\mathrm{Al})$, i.e. $n=\frac{w}{M}=\frac{54}{27}=2 \mathrm{~mol}$
Heat required $(Q)=n C_s \Delta T$
$$
=2 \times 24 \times 20=960 \mathrm{~J}
$$
Hence, option (c) is the correct answer.
Weight of aluminium $(w)=54 \mathrm{~g}$
Temperature difference $(\Delta T)=60^{\circ}-40^{\circ}=20^{\circ} \mathrm{C}$
Molar heat capacity of aluminium
$$
\left(C_s\right)=24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
$$
Atomic mass of $(\mathrm{Al})$, i.e. $(M)=27$
$\because \operatorname{Moles}$ of $(\mathrm{Al})$, i.e. $n=\frac{w}{M}=\frac{54}{27}=2 \mathrm{~mol}$
Heat required $(Q)=n C_s \Delta T$
$$
=2 \times 24 \times 20=960 \mathrm{~J}
$$
Hence, option (c) is the correct answer.
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