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A metal ball of mass 0.1 kg is heated upto $500^{0} C$ and dropped into a vessel of heat capacity $800 J K^{- 1}$ and containing $0.5 k g$ water. The initial temperature of water and vessel is $30^{0} C$ . 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 $400 J k g^{- 1} K^{- 1}$ ]

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answer is 21.

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Detailed Solution

Assume final temperature $= T^{0} C$
Heat less = Heat gain = $m s Δ T$
$\Rightarrow m_{B} s_{B} Δ T_{B} = m_{w} s_{w} Δ T_{w}$

$0.1 \times 400 \times (500 - T) = 0.5 \times 4200 \times (T - 30) + 800 (T - 30)$

$\Rightarrow 40 (500 - T) = (T - 30) (2100 + 800) \Rightarrow 2000 - 40 T = 2900 T - 30 \times 2900$

$\Rightarrow 2000 + 30 \times 2900 = T (2940) \Rightarrow T = {30.4}^{0} C \frac{Δ T}{T} \times 100 = \frac{6.4}{30} \times 100 = 21 %$

So the closest answer is 20%.

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