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Standard XII

Chemistry

Stoichiometric Calculations

A beaker cont...

Question

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

58^{o} C

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68^{o} C

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73^{o} C

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78^{o} C

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Solution

The correct option is A $68^{o} C$
Assuming that the beaker is also at $20^{o}$ C
Heat lost by the hot water=heat gained by the water+heat gained by the beaker
$440. C . (92 - T) = 200. C . (T - 20) + 20. C . (T - 20)$ (C is the specific heat capacity of water)
$440 (92 - T) = 220 (T - 20)$
$184 - 2 T = T - 20$
$T = 68^{o} C$

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

The correct option is A 68oCAssuming that the beaker is also at 20oCHeat lost by the hot water=heat gained by the water+heat gained by the beaker440.C.(92−T)=200.C.(T−20)+20.C.(T−20) (C is the specific heat capacity of water)440(92−T)=220(T−20)184−2T=T−20T=68oC

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