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A vessel contains 110 g of water. The heat capacity of the vessel is equal to 10 g of water. The initial temperature of water in vessel is 10°C. If 220 g of hot water at 70°C is poured in the vessel, the final temperature neglecting radiation loss, will be

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$70^{0} C$

$80^{0} C$

$50^{0} c$

$60^{0} C$

answer is D.

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Detailed Solution

Let final temperature of water be $θ$
Heat taken = Heat given

$110 \times 1 \times (θ - 10) + 10 (θ - 10) = 220 \times 1 \times (70 - θ)$

$\Rightarrow \Rightarrow θ = {48.8}^{\circ} C ≃ 50^{\circ} C$ $θ$ = 48.8°C = 50°C

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