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A metal ball of mass $0.1 \mathrm{~kg}$ is heated upto $500^{\circ} \mathrm{C}$ and dropped into a vessel of heat capacity $800 \mathrm{JK}^{-1}$ and containing $0.5 \mathrm{~kg}$ water. The initial temperature of water and vessel is $30^{\circ} \mathrm{C}$. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, $4200 \mathrm{Jkg}^{-}$ ${ }^{1} \mathrm{~K}^{-1}$ and $400 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ ]
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Verified Answer
The correct answer is:
$20 \%$
Assume final temperature $=\mathrm{T}^{\circ} \mathrm{C}$ Heat lass $=$ Heat gain $=\mathrm{ms} \Delta \mathrm{T}$ $\Rightarrow \mathrm{m}_{\mathrm{B}} \mathrm{s}_{\mathrm{B}} \Delta \mathrm{T}_{\mathrm{B}}=\mathrm{m}_{\mathrm{w}} \mathrm{s}_{\mathrm{w}} \Delta \mathrm{T}_{\mathrm{w}}$
$0.1 \times 400 \times(500-\mathrm{T})$
$\quad=0.5 \times 4200 \times(\mathrm{T}-30)+800(\mathrm{~T}-30)$
$\Rightarrow 40(500-\mathrm{T})=(\mathrm{T}-30)(2100+800)$
$\Rightarrow 20000-40 \mathrm{~T}=2900 \mathrm{~T}-30 \times 2900$
$\Rightarrow 20000+30 \times 2900=\mathrm{T}(2940)$
$\mathrm{T}=30.4^{\circ} \mathrm{C}$
$\frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{6.4}{30} \times 100=21 \%$
so the closest answer is $20 \%$.
$0.1 \times 400 \times(500-\mathrm{T})$
$\quad=0.5 \times 4200 \times(\mathrm{T}-30)+800(\mathrm{~T}-30)$
$\Rightarrow 40(500-\mathrm{T})=(\mathrm{T}-30)(2100+800)$
$\Rightarrow 20000-40 \mathrm{~T}=2900 \mathrm{~T}-30 \times 2900$
$\Rightarrow 20000+30 \times 2900=\mathrm{T}(2940)$
$\mathrm{T}=30.4^{\circ} \mathrm{C}$
$\frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{6.4}{30} \times 100=21 \%$
so the closest answer is $20 \%$.
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