A hot water cools from `92^(@)C` to `84^(@)C` in 3 minutes when the room temperature is `27^(@)C`. How long will it take for it to cool from `65^(@)C` to` 60^(@)C`?

Explanation

As the temperature difference are small, be can use newton's law of cooling.
`(d theta)/(dt)=-k(theta-theta_(0))` . Or, `(d theta)/(theta-theta_(0))=-kdt` . Where k is a constance, theta is the temperature of the body at time t and `theta_(0)=16^(@)C` . is the temperature of the surrounding. We have,
`int_(40^(@)C)^(36^(@)C)(d theta)/(theta-theta_(0))=-k(5min)` . or, `ln`(36^(@)C-16^(@)C)/(40^(@)C-16^(@)C)=-k(5min)` . or, `k=-ln(5//6)/(5mint)` . If t be the time required for the tempearature to fall from `36^(@)C` to `32^(@)C` then by (i),
`int_(36^(@)C)^(32^(@)C)(d theta)/(theta-theta_(0))=-kt` . or, `ln`(32^(@)C-16^(@)C)/(36^(@)C-16^(@)C)=(ln(5//6)t)/(5mint)` . or, `t=ln(4//5)/(ln(5//6))xx5min` . `=6.1min` .
Alternative method The mean temperature of the body as it cools from `40^(@)C` to `36^(@)C` is `(40^(@)C+36^(@)C)/(2)=38^(@)C` . The rate of decrease of temperature is `(40^(@)C+36^(@)C)/(5min)=0.80^(@)C` min^(-1)` . Netton's law o fcooling is `(d theta)/(dt)=-k(theta-theta_(0))` . or, `-0.8^(@)Cmin^(-1)=-k,(38^(@)C-16^(@)C)=-k(22^(@)C)` . or, k=(0.8)/(22)min^(-1)` . Let the time taken for the temperature to become `32^(@)C` . be t. During this period,
`(d theta)/(dt)=-(36^(@)C-32^(@)C)/(t)=-(4^(@)C)/(t)` . The mean temperature is `(36^(@)C+32^(@)C)/(2)=34^(@)C` . Now, (d theta)/(dt)=-k(theta-theta_(0)` . or, `(-4^(@)C)/(t)=-(8.0)/(22)xx(34^(@)C-16^(@)C)min^(-1)` . or, `t=(22xx4)/(0.8xx18)min=6.1min` .