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PhysicsHeat & ThermoRadiationMedium2 minQB

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Passage / Comprehension

A body cools in surroundings at a constant 30°C. Its heat capacity is 2 J/°C and its initial temperature is 40°C. Assume Newton's law of cooling. It cools to 36°C in 10 minutes.

The temperature of the body in °C denoted by $θ$ the variation of $θ$ versus time t is best denoted as

Options:

Answer:

Solution:

$\frac{d θ}{d t} = - k (θ - 3 0^{\circ} C)$
As the body cools, the magnitude of the slope decreases and the temperature approaches 30°C.
Answer D.

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Radiation

⏱ 2mℹ️ Source: QB

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