Question:
The rate at which a substance cools in moving air, is proportional to the difference between the temperature of the substance and that of air. The temperature of air is 290 K and the substance cools from 370 K to 330 K in 10 minutes. Then the time to cool the substance upto 295 K is
The rate at which a substance cools in moving air, is proportional to the difference between the temperature of the substance and that of air. The temperature of air is 290 K and the substance cools from 370 K to 330 K in 10 minutes. Then the time to cool the substance upto 295 K is
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Write the cooling law as dT/dt = -k(T - 290), solve for T, and use the two conditions to find k and the time.
Updated On: Nov 13, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Newton's law gives \(\dfrac{dT}{dt}=-k(T-290)\), so \(T-290=Ce^{-kt}\).
Step 2: Find the constants
At \(t=0\), \(T=370\), so \(C=80\).
At \(t=10\), \(T=330\): \(40=80e^{-10k}\), so \(e^{-10k}=\tfrac12\).
Step 3: Time for 295 K
\(295-290=5=80e^{-kt}\), so \(e^{-kt}=\tfrac1{16}=\left(\tfrac12\right)^4\).
Hence \(kt=4\times10k\), so \(t=40\) min.
Final Answer:
The substance cools to 295 K in 40 minutes. \[ \boxed{\text{(A) }40\text{ min}} \]
Newton's law gives \(\dfrac{dT}{dt}=-k(T-290)\), so \(T-290=Ce^{-kt}\).
Step 2: Find the constants
At \(t=0\), \(T=370\), so \(C=80\).
At \(t=10\), \(T=330\): \(40=80e^{-10k}\), so \(e^{-10k}=\tfrac12\).
Step 3: Time for 295 K
\(295-290=5=80e^{-kt}\), so \(e^{-kt}=\tfrac1{16}=\left(\tfrac12\right)^4\).
Hence \(kt=4\times10k\), so \(t=40\) min.
Final Answer:
The substance cools to 295 K in 40 minutes. \[ \boxed{\text{(A) }40\text{ min}} \]
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