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A heated irron block at 127^(@)C loses 3...

A heated irron block at `127^(@)C` loses `300J` of heat to the surroundings which are at a temperature of `27^(@)C`. This process is `0.05 J K^(-1)`. Find the value of `x`.

Text Solution

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`Delta_(sys)S = (q_(sys))/(T_(sys)) =- (300)/(273+127)`
`= (-300)/(400) (-3)/(4) J K^(-1)`
`Delta_(surr)S = (-q_(sys))/(T_(surr)) =- (300)/(273+27)`
`= (300)/(300) = + 1JK^(-1)`
`Delta_(total)S or Delta_("universe")S = Delta_(sys)S + Delta_(surr)S`
`= (-3)/(4) +1 = (1)/(4) = 0.25 J K^(-1)`
`:. 0.05x = 0.25`
`x = 5`
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