Hello Forumites,

A cup of coffee, cooling off in a room at a temperature $20^{\circ}C,$ has cooling constant $k=0.09min^{-1}$. Now a)I want to know how fast is the coffee cooling(in degrees per minute) when its temperature is $T=80^{\circ}C?$

b) I want to use linear approximation to estimate the change in temperature over the next 6 seconds when $T=80^{\circ}C$

c) The coffee is served at a temperature of $90^{\circ}C.$ How long should I wait before drinking it if the optimal temperature is $65^{\circ}C$?

Answer:-a)$ \frac{dT}{dt}|_{T=80^{\circ}C}=-0.09*(80-20)=-5.4^{\circ}C/min.$

b) Note that 6 seconds should be used as 0.1 minutes.From $T\approx 80-5.4\Delta t=80-5.4*0.1$. It follows that the change of temperature will be $T-80\approx -0.54^{\circ}C$

c) How to answer c)? Answer provided to me is 6.4 minutes approx.