Ordinary Differential Equations Questions and Answers – Newton’s Law of Cooling and Escape Velocity

This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Newton’s Law of Cooling and Escape Velocity”.

1. 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) ekt
d) T=t2+(t1) ekt

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=ec=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.

2. A body in air at 25℃ cools from 100℃ to 75℃ in 1 minute. What is the temperature of the body at the end of 3 minutes? (Take log(1.5)=0.4)
a) 40℃
b) 47.5℃
c) 42.5℃
d) 50℃

Explanation: By Newton’s law of cooling w.k.t T = t2 + (t1-t2) e-kt, given t1=100℃, t2=25℃
when t=1 –> T(1) = 25 + 75 e-k = 75℃ –> 50/75 = 2/3 = e-k
–> 3/2=ek taking log k=log(1.5)=0.4.
to find T when t=3 minute using the value of k we get
T = 25 + 75e-0.4*3 = 47.5℃……e-1.2=0.3.

3. A bottle of mineral water at a room temperature of 72℉ is kept in a refrigerator where the temperature is 44℉.After half an hour water cooled to 61℉.What is the temperature of the body in another half an hour?(Take log $$\frac{28}{17}$$ = 0.498, e-0.99=0.37)
a) 18℉
b) 9.4℉
c) 54.4℉
d) 36.4℉

Explanation: By Newton’s law of cooling w.k.t T=t2+(t1-t2) e-kt, given t1=72℉, t2=44℉
At t=half an hour = 30mts T=61℉, finding k using the given values i.e
61=44+28e-k30 –> $$\frac{17}{28}$$ = e-k30 or $$\frac{28}{17}$$ = ek30 taking log, log $$\frac{28}{17}$$ = 30k –> k=0.0166
to find T when t = 30mts + 30mts = 60mts
T = 44 + 28e-(0.0166)30 = 54.4℉.

4. The radius of the moon is roughly 2000km. The acceleration of gravity at the surface of the moon is about $$\frac{g}{6}$$, where g is the acceleration of gravity at the surface of the earth. What is the velocity of escape for the moon?(Take g=10ms-2)
a) 2.58 kms-1
b) 4.58 kms-1
c) 6.28 kms-1
d) 12.28 kms-1

Explanation: Let R be radius of the earth and r be the variable distance from Newton’s law $$a = \frac{dv}{dt} = \frac{k}{r^2}$$ when r=R a=-g due to retardation of a body -gR2=k
substituting the value of k back we get $$\frac{dv}{dt} = \frac{-gR^2}{r^2} = \frac{dr}{dt} \frac{dv}{dr} = v \frac{dv}{dr}$$ solving DE for v
we get $$\int v \,dv = \int \frac{-gR^2}{r^2} dr + c \rightarrow v^2 = \frac{2gR^2}{r} + C$$ to find c we use at r=R
v=ve thus we get $$v_e^2 – \frac{2gR^2}{R} = C$$… where 2c=C=constant substituting value of c we get $$v^2=\frac{2gR^2}{r} + v_e^2 – 2gR$$…..if r>>R $$\frac{2gR^2}{r} = 0$$ and particle to get escape from earth v≥0 –> ve2 – 2gR≥0 –> $$v_e=\sqrt{2gR}$$ to find ve
from the moon g becomes $$\frac{g}{6}$$, R=2000km=2×106 m, g=10ms-2
therefore $$v_e = \sqrt{2*\frac{10}{6}(2×10^6)} = 2.58kms^{-1}.$$.

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