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A liquid takes 5 minutes to cool from `80^@C` to` 50^@C` . How much time will it take to cool from `60^@C` to `30^@C`? The temperature of surroundings is `20^@C`. |
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Answer» Let us start with the assumption that `|(Deltatheta)/Delta| prop "Temperature difference"` `:. ((80-50)/5) = alpha [(80+50)/2 -20]` and `(60-30)/t = alpha [(60+30)/2 -20]` Solving these two equations, we get `t = 9 min` . |
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