A hot cup of local tea prepared for a nursing mother cools from 96 degrees Celsius to 60 degree Celsius after 240 seconds, find the time it will take for the tea to cool to a drinkable temperature of 45 degrees Celsius?

1 Answer
A08
Apr 10, 2018

We know that solution of differential equation of Newton's Law of Cooling is

#T(t)=T_a+(T_0-T_a)e^(-kt)#
where #T(t)# is temperature of sample as a function of time, #T_a# is ambient temperature, #T_0# is initial temperature and #k# is a positive constant.

Taking ambient temperature as #20^@C# and inserting given values in the equation we get

#60=20+(96-20)e^(-240k)#
#=>76e^(-240k)=40#
#=>-240k=ln(40/76)#
#=>k=ln(76/40)/240#
#=>k=ln(76/40)/240#
#=>k=0.00267\ "per sec"#

Equation becomes

#T(t)=20+76e^(-0.00267t)#

Time taken to cool the tea to drinkable #45^@C#

#45=20+76e^(-0.00267t)#
#=>e^(-0.00267t)=25/76#
#=>t=ln(76/25)/0.00267#
#=>t=416\ s#, rounded to units place.

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