A liquid takes `10` minutes to cool from `80^(@)C` to `50^(@)C`. The temperature of the surroudings is `20^(@)C`. Assuming that the Newton's law of cooling is obeyed, the cooling constant will be -

A hot liquid takes 10 minute to cool from 70^(@)C to 60^(@)C . The time taken by the liquid to cool from 60^(@)C to 50^(@)C is

A body cools from 50.0^(@)C to 49.9^(@)C in 10s. How long will it take to cool from 40.0^(@)C to 39.9^(@)C ? Assume the temperature of the surrounding to be 30.0^(@)C and Newton's law of cooling to be valid.

A body cools from 50.0^(@)C to 49.9^(@)C in 5s. How long will it take to cool from 40.0^(@)C to 39.9^(@)C ? Assume the temperature of the surroundings to be 30.0^(@)C and Newton's law of cooling to be valid:

A liquid contained in a beaker cools from 80^(@)C to 60^(@)C in 10 minutes. If the temperature of the surrounding is 30^(@)C , find the time it will take to cool further to 48^(@)C . Here cooling take place ac cording to Newton's law of cooling.

A body cools in 7 minutes from 60^(@)C to 40^(@)C . What will be its temperature after the next 7 minutes? The temperature of the surrounding is 10^(@)C . Assume that Newton's law of cooling holds good throughout the process.

A body cools from 50.0^(@)C to 48^(@)C in 5s. How long will it take to cool from 40.0^(@)C to 39^(@)C ? Assume the temperature of surroundings to be 30.0^(@)C and Newton's law of cooling to be valid.

A body cool from 50.0^(@) C to 39.9^(@) C in 5 s . How long will it take to cool from 40.0^(@) C to 39.9^(@) C ? Assume the temperature of surroundings to be 30^(@) C and Newton's law of cooling to be valid