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Standard X

Physics

Calorimeter

A beaker cont...

Question

A beaker contains $200$ g of water.The heat capacity of beaker is equal to that $20$ g of water.The initial temperature of water in the beaker is $20^{\circ} C$ . If $440$ g of hot water at $92^{\circ} C$ is poured in the final temperature,neglecting radiation loss, will be

58^{\circ} C

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68^{\circ} C

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73^{\circ} C

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78^{\circ} C

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Solution

The correct option is A $68^{\circ} C$
Let the final temperature is $T^{o} C$
Heat lost by hot body= Heat gained by cold body
$m_{1} s_{1} Δ T_{1} = m_{2} s_{2} Δ T_{2}$
Solving it in CGS units,
$200 \times 1 \times (T - 20) + 20 \times 1 \times (T - 20) = 440 \times 1 \times (92 - T)$
$T = 68^{o} C$

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A beaker contains 200 g of water.The heat capacity of beaker is equal to that 20 g of water.The initial temperature of water in the beaker is 20∘C. If 440 g of hot water at 92∘C is poured in the final temperature,neglecting radiation loss, will be

The correct option is A 68∘CLet the final temperature is ToCHeat lost by hot body= Heat gained by cold bodym1s1ΔT1=m2s2ΔT2Solving it in CGS units,200×1×(T−20)+20×1×(T−20)=440×1×(92−T)T=68oC

A beaker contains $200$ g of water.The heat capacity of beaker is equal to that $20$ g of water.The initial temperature of water in the beaker is $20^{\circ} C$ . If $440$ g of hot water at $92^{\circ} C$ is poured in the final temperature,neglecting radiation loss, will be