Question

Two identical metal bars are heated in two different temperatures and allowed to cool in the same surroundings. Which one of the following figures correctly shows their cooling curves?

• A. Fig A
• B. Fig B
• C. Fig C
• D. Fig D

Hint:

Cooling curves for identical objects in the same surroundings can never intersect. At any given temperature, the instantaneous rate of cooling is uniquely determined, so the curves must remain parallel-like, gradually converging at infinity.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the surrounding temperature.
Mathematically, \(\frac{dT}{dt} = -k(T - T_s)\), where \(T_s\) is the surrounding temperature.
This differential equation leads to an exponential decay of temperature over time.
Two identical bars will have the same cooling constant \(k\).
Step 2: Key Formula or Approach:
The solution to the cooling equation is:
\[ T(t) = T_s + (T_0 - T_s) e^{-kt} \]
where \(T_0\) is the initial temperature.
Since the bars are identical and in the same surroundings, \(k\) and \(T_s\) are the same for both.
Step 3: Detailed Explanation:
Consider two initial temperatures \(T_{01}\) and \(T_{02}\) where \(T_{01}>T_{02}\).
The equations for temperature as a function of time are:
\[ T_1(t) = T_s + (T_{01} - T_s) e^{-kt} \]
\[ T_2(t) = T_s + (T_{02} - T_s) e^{-kt} \]
Since \(T_{01} - T_s>T_{02} - T_s\), at any time \(t\), the term \((T_{01} - T_s) e^{-kt}\) will be strictly greater than \((T_{02} - T_s) e^{-kt}\).
Consequently, \(T_1(t)\) will always be greater than \(T_2(t)\).
The curves will never cross each other.
Both curves will asymptotically approach the same surrounding temperature \(T_s\) as \(t \to \infty\).
The rate of cooling (slope) is higher for the bar at a higher temperature.
Graph (D) correctly shows two exponential decay curves that start at different temperatures and approach the same baseline without crossing.
Step 4: Final Answer:
Newton's law of cooling predicts exponential decay towards the ambient temperature. For identical bars, the curves are parallel in behavior and never intersect, as shown in figure (D).