Kinetic Theory

JEE Main level, single correct, medium

106 of 168

Question 106 of 168

A metal ball of mass 0.1 kg0.1 \mathrm{~kg} is heated upto 500∘C500^{\circ} \mathrm{C} and dropped into a vessel of heat capacity 800JK−1800 \mathrm{JK}^{-1} and containing 0.5 kg0.5 \mathrm{~kg} water. The initial temperature of water and vessel is 30∘C30^{\circ} \mathrm{C}. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200Jkg−4200 \mathrm{Jkg}^{-} 1 K−1{ }^{1} \mathrm{~K}^{-1} and 400Jkg−1 K−1400 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1} ]

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Answer: (D) 20%20 \%

Assume final temperature =T∘C=\mathrm{T}^{\circ} \mathrm{C} Heat lass == Heat gain =msΔT=\mathrm{ms} \Delta \mathrm{T} ⇒mBsBΔTB=mwswΔTw\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}}

0.1×400×(500−T)0.1 \times 400 \times(500-\mathrm{T})

=0.5×4200×(T−30)+800( T−30)\quad=0.5 \times 4200 \times(\mathrm{T}-30)+800(\mathrm{~T}-30)

⇒40(500−T)=(T−30)(2100+800)\Rightarrow 40(500-\mathrm{T})=(\mathrm{T}-30)(2100+800)

⇒20000−40 T=2900 T−30×2900\Rightarrow 20000-40 \mathrm{~T}=2900 \mathrm{~T}-30 \times 2900

⇒20000+30×2900=T(2940)\Rightarrow 20000+30 \times 2900=\mathrm{T}(2940)

T=30.4∘C\mathrm{T}=30.4^{\circ} \mathrm{C}

ΔTT×100=6.430×100=21%\frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{6.4}{30} \times 100=21 \%

so the closest answer is 20%20 \%.

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