Question

A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

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Solution

According space to space Newton ’ straight s space law space of space cooling comma fraction numerator negative space dT over denominator straight T end fraction space equals space straight K space left parenthesis straight T space minus space straight T subscript straight o right parenthesis space fraction numerator dT over denominator straight K thin space left parenthesis straight T space minus space straight T subscript straight o right parenthesis end fraction space equals space minus space Kdt space space space space space space space space space space space space space space... space left parenthesis straight i right parenthesis space

where,
T is the temperature of the body,
Temperature of the surroundings, To = 20⁰ C, and
K is a constant.
Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s

Now comma space integrating space equation space left parenthesis straight i right parenthesis comma space we space get space space space space space space space space integral subscript 50 superscript 80 fraction numerator dT over denominator straight K left parenthesis straight T space minus space straight T subscript straight o right parenthesis end fraction space equals space minus space integral subscript 0 superscript 300 straight K. space dt space space space space space space space space space space left square bracket log subscript straight e left parenthesis straight T space minus space straight T subscript straight o right parenthesis right square bracket subscript 50 superscript 80 space equals space minus straight K space left square bracket straight t right square bracket space subscript 0 superscript 300 space rightwards double arrow fraction numerator 2.3026 over denominator straight K end fraction space log subscript 10 space 2 space space equals space minus 300 space rightwards double arrow fraction numerator negative space 2.3026 over denominator 300 end fraction space log subscript 10 space 2 space equals space straight K space space space space space space space space space space space space space... space left parenthesis ii right parenthesis Temperature space of space the space body space falls space from space 60 to the power of straight o space straight C space to space 30 to the power of straight o space straight C space in space time space straight t apostrophe. Hence comma space we space have space space space space space space space fraction numerator 2.3026 over denominator straight K space end fraction space log subscript 10 space fraction numerator 60 minus 20 over denominator 30 space minus space 20 end fraction space equals space minus straight t to the power of 1 space rightwards double arrow space space minus fraction numerator 2.3026 over denominator straight t apostrophe end fraction space log subscript 10 space 4 space equals space straight K space space space space space space space space space space... space left parenthesis iii right parenthesis On space equating space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis comma space we space get space minus fraction numerator 2.3026 over denominator straight t apostrophe end fraction space log subscript 10 space 4 space equals space fraction numerator negative 2.3026 over denominator 300 end fraction space log subscript 10 space 2 space rightwards double arrow space straight t apostrophe space equals space 300 cross times 2 space equals space 600 space straight s space equals space 10 space min Therefore comma space 10 space minutes space is space taken space to space cool space the space body space from space 60 to the power of straight o space straight C space to space 30 to the power of straight o space straight C.

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Mechanical Properties Of Fluids

A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

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