NEET Physics Newton's Law of Cooling MCQ Question
A body is placed in an environment where the surrounding temperature is 20°C. Initially, the body is at 100°C. If the cooling constant is 0.1 min⁻¹, what is the rate of change of temperature when the body has cooled to 60°C?
-4.0°C/min
-8.0°C/min
-2.0°C/min
-6.0°C/min
Correct Answer
Detailed Explanation
According to Newton's Law of Cooling, the rate of cooling is proportional to the temperature difference between the body and the surroundings, given by dQ/dt = -k(T2 - T1). For T2 = 60°C and T1 = 20°C, dQ/dt = -0.1(60 - 20) = -4.0°C/min.
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