5 kg of water at 10-C is added to 10 kg of water at 40-C Neglecting heat capacity or vessel and other losses- the equilibrium temperature will be

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Standard X

Physics

Calorimeter

5 kg of water...

Question

$5$ kg of water at $10^{\circ} C$ is added to $10$ kg of water at $40^{\circ} C$ Neglecting heat capacity or vessel and other losses, the equilibrium temperature will be

30^{\circ} C

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25^{\circ} C

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35^{\circ} C

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33^{\circ} C

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Solution

The correct option is C $30^{\circ} C$
Heat gained by water at $10^{o} C$ =heat lost by water at $40^{o} C$
$T$ is the final temperature.
so, $5 \times c \times (T - 10) = 10 \times c \times (40 - T)$
$\Rightarrow T = 30^{o} C$

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5 kg of water at 10∘C is added to 10 kg of water at 40∘C Neglecting heat capacity or vessel and other losses, the equilibrium temperature will be

The correct option is C 30∘CHeat gained by water at 10oC=heat lost by water at 40oCT is the final temperature.so, 5×c×(T−10)=10×c×(40−T)⇒T=30oC

$5$ kg of water at $10^{\circ} C$ is added to $10$ kg of water at $40^{\circ} C$ Neglecting heat capacity or vessel and other losses, the equilibrium temperature will be

The correct option is C $30^{\circ} C$
Heat gained by water at $10^{o} C$ =heat lost by water at $40^{o} C$
$T$ is the final temperature.
so, $5 \times c \times (T - 10) = 10 \times c \times (40 - T)$
$\Rightarrow T = 30^{o} C$