A body cools from `70^(@)C` to `50^(@)C` in 6 minutes when the temperature of the surrounding is `30^(@)C`. What will be the temperature of the body after further 12 minutes if coolilng proseeds ac cording to Newton's law of cooling ?

A body cools from 60^(@) C to 50^(@) C in 6 minutes, the temperature of the surroundings being 25^(@) C. What will be its temperature after another 6 minutes ?

The temperature of a body falls from 50^(@)C to 40^(@)C in 10 minutes. If the temperature of the surroundings is 20^(@)C Then temperature of the body after another 10 minutes will be

The temperature of a body falls from 50^(@)C to 40^(@)C in 10 minutes. If the temperature of the surroundings is 20^(@)C Then temperature of the body after another 10 minutes will be

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 from 70^(@)C to 50^(@)C in 5 minutes. Temperature of surroundings is 20^(@)C . Its temperature after next 10 minutes is

A body cools from 70^(@)C to 50^(@)C in 5minutes Temperature of surroundings is 20^(@)C Its temperature after next 10 minutes is .

A body cools from 71^(@) C. to 50^(@) C in 5 minutes. Temperature of surroundings is 20^(@) C. Its temperature after next 10 minutes is

A body initially at 64^(@)C cools to 52^(@)C in 5 minutes. The temperature of surroundings is 16^(@)C . Find the temperature after furthre 5 minutes.

A body takes 10 minutes to cool from 60^(@)C to 50^(@)C . The temperature of surroundings is constant at 25^(@) C. Then, the temperature of the body after next 10 minutes will be approximately

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