A hot coffee at 80 c is left in a room at 20 c.it is found that after 20 minutes in the room, the temperature of coffee has decreased by 20 c. Determine the temperature of coffee after 30 minutes?

1 Answer
Mar 16, 2018

I get #53^@"C"# after #30# minutes have passed.

Explanation:

We use Newton's Law of Cooling, which states that

#(dT)/(dt)=K(T-T_0)#

Solving the differential equation, we get #T=T_0+Ae^(Kt)#.

where #t# is the time in minutes, #T_0# is the surrounding temperature, #T# is the object temperature, and #K# is the constant (varies).

Plugging in the values, we need to find #K# and #A# first.

We got:

#T_0=20#

#:.T=20+Ae^(Kt)#

When #t=0,T=80#, and so

#80=20+Ae^0#

#80=20+A#

We get:

#A=60#

The equation becomes:

#:.T=20+60e^(Kt)#

Now, we can find #K#.

Decreasing by #20^@"C"#, the new temperature is therefore #80-20=60^@"C"#.

#60=20+60e^(20K)#

#e^(20K)=40/60#

#=2/3#

#20K=ln(2/3)#

#K=(ln(2/3))/20#

#~~-0.02#

So, the final equation is:

#T=20+60e^(-0.02t)#

After #30# minutes, we plug in #t=30#, and we get

#T=20+60e^(-0.02*30)#

#=20+60e^-0.6#

#=20+32.9#

#=52.9#

#~~53^@"C"#

So, the temperature after half an hour will be approximately #53# degrees celsius.