A metal ball of mass $0.1$ kg is heated upto $500^{o}$ C and dropped into a vessel of heat capacity $800$ $J K^{- 1}$ and containing $0.5$ kg water. The initial temperature of water and vessel is $30^{o}$ C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, $4200$ $J k g^{- 1} K^{- 1}$ and $400$ $J K g^{- 1} K^{- 1}$ ]

30 %

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20 %

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25 %

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15 %

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Solution

The correct option is A $20 %$
$0.1 \times 400 \times (500 - T) = 0.5 \times 4200 \times (T - 30) + 800 (T - 30)$
$\Rightarrow 40 (500 - T) = (T - 30) (2100 + 800)$
$\Rightarrow 20000 - 40 T = 2900 T - 30 \times 2900$
$\Rightarrow 20000 + 30 \times 2900 = T (2940)$
$T = {30.4}^{o} C$
$\frac{Δ T}{T} \times 100 = \frac{6.4}{30} \times 100$
$= 20 %$ .

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