According to the Newton's law of cooling, the rate of change of temperature of a hot object at any time t is proportional to the difference between the temperature of the object an... According to the Newton's law of cooling, the rate of change of temperature of a hot object at any time t is proportional to the difference between the temperature of the object and the temperature of the surrounding at that time. Assuming this law, solve the following problem: An iron rod is heated to 400°C and cooled in a room with fixed temperature of 25°C. After 4 minutes, the temperature of the rod drops down to A°C. What will be its temperature after 8 minutes? Read more

Steps to Solve

1. Identify the given information The initial temperature of the iron rod, $T_0$, is $400^\circ C$. The surrounding temperature, $T_s$, is $25^\circ C$. After $4$ minutes, the temperature drops to $A^\circ C$.

2. State Newton's Law of Cooling According to Newton's Law of Cooling, the rate of change of temperature is given by: $$ \frac{dT}{dt} = -k(T - T_s) $$ Where:

• $T$ is the temperature of the object
• $T_s$ is the surrounding temperature
• $k$ is the constant of proportionality

3. Set up the differential equation Using the temperatures given, we can write the equation specifically for the iron rod: $$ \frac{dT}{dt} = -k(T - 25) $$

4. Separate variables Rearranging gives: $$ \frac{1}{T - 25} dT = -k dt $$

5. Integrate both sides Integrate the left side and the right side: $$ \int \frac{1}{T - 25} dT = -k \int dt $$ This results in: $$ \ln |T - 25| = -kt + C $$

6. Solve for T Exponentiating both sides gives: $$ |T - 25| = e^{-kt + C} $$ Letting $C_1 = e^C$, we have: $$ T - 25 = C_1 e^{-kt} $$

7. Find constant C_1 At $t = 0$, $T = 400$: $$ 400 - 25 = C_1 e^{0} $$ So, $$ C_1 = 375 $$

8. Write the equation for T The equation for $T$ becomes: $$ T - 25 = 375 e^{-kt} $$ or $$ T(t) = 375 e^{-kt} + 25 $$

9. Determine the rate constant k We know after $4$ minutes, the temperature dropped to $A$. Substitute $t = 4$: $$ A - 25 = 375 e^{-4k} $$

10. Use the temperature after 4 minutes to find k Isolate $e^{-4k}$: $$ e^{-4k} = \frac{A - 25}{375} $$

11. Calculate the temperature after 8 minutes Substituting back into the equation for $T$, but now for $t = 8$: $$ T(8) = 375 e^{-8k} + 25 $$

12. Substituting $e^{-4k}$ Since $e^{-8k} = (e^{-4k})^2$, we can substitute: $$ T(8) = 375 \left(\frac{A - 25}{375}\right)^2 + 25 $$

13. Final Formula Substitution This gives: $$ T(8) = 375 \frac{(A - 25)^2}{375^2} + 25 $$