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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`. |
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Answer» `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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