Question

According to Newton’s law of cooling “The change of temperature of a body is proportional to the difference between the temperature of a body and that of the surrounding medium”. If t1℃ is the initial temperature of the body and t2℃ is the constant temperature of the medium, T℃ be the temperature of the body at any time t then find the expression for T℃ as a function of t1℃, t2℃ and time t.

(a) T=t1+(t2) e^-kt

(b) T=t2+(t1-t2) e^-kt

(c) T=t1+(t1-t2) e^kt

(d) T=t2+(t1) e^kt

I got this question in semester exam.

This key question is from Newton’s Law of Cooling and Escape Velocity topic in section Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right answer is (b) T=t2+(t1-t2) e^-kt

The best explanation: According to the definition of Newton’s law of cooling \(\frac{dT}{dt} ∝ (T-t_2) \,or\, \frac{dT}{dt} = -k(T-t_2)\) ….k is a constant of proportionality and negative sign indicates the cooling of a body with increase of the time. since t1℃ initial temperature of the body at t=0 T=t1 –> T(0) = t1℃. \(\frac{dT}{dt} = -k(T-t_2)\)…….at T(0) = t1℃, now solving DE the above equation is of variable separable form i.e \(\int \frac{dt}{T-t_2} = \int -kdt + c\)

=log (T-t2) = -kt + c –> T-t2 = pe^-kt…where p=e^c=constant, using initial condition i.e T(0)= t1 we get t1-t2=p substituting back in equation we obtain  T=t2+(t1-t2) e^-kt.