A thermometer is taken from a room where the temperature is 23 ^o C to the outdoors, where the temperature is -15 ^o C. After one minute the thermometer reads 11 ^o C. ?

A thermometer is taken from a room where the temperature is 23 ^o C to the outdoors, where the temperature is -15 ^o C. After one minute the thermometer reads 11 ^o C.
(a) What will the reading on the thermometer be after 5 more minutes?

,
(b) When will the thermometer read -14 ^o C?

minutes after it was taken to the outdoors.

1 Answer
Jan 26, 2018

Answer:

(a) - #-11.1^@C#
(b) - after #575# seconds i.e. #8# minutes and #35# seconds temperature would be #-14^@C#

Explanation:

According to Newton's Law of Cooling, the rate of the cooling is proportional to the temperature difference between the object and its surroundings.

In other words, if an object at a higher temperature say #T_0# is moved to a surrounding at a lower temperature #T_s#, the rate of cooling is directly proportional to the difference in temperature of te two i.e. #(dT)/(dt)=-k(T_0-T_s)#. This means that temperature #T_t# of the object at #t# time is given by #T_t=T_s+(T_0-T_s)e^(-kt)#

Here we have #T_0=23^@#, #T_s=-15^@# and at #t=60#, we have #T_60=11^@# - here we have used #60# as we intend to use time in seconds.

Hence #11=-15+(23-(-15))e^(-60k)# or #26=38e^(-60k)#

or #e^(-60k)=26/38# and hence

#k=-ln(26/38)/60=0.00632482702#

(a) After #5# more minutes i.e. after #360# seconds, temperature would be

#T_360=-15+38e^(-360xx0.00632482702)#

= #-15+3.9=-11.1^@#

(b) Let the temperature be #-14^@C# after #t# seconds, then

#-14=-15+38e^(-0.00632482702t)#

or #e^(-0.00632482702t)=(15-14)/38=0.026315789#

i.e. #t=-ln0.026315789/0.00632482702~=575#

Hence after #575# seconds i.e. #8# minutes and #35# seconds temperature would be #-14^@C#