Question

A metal ball cools at 2.0°C/min when its temperature is 60°C and the surrounding temperature is 20°C. What will be its rate of cooling when its temperature drops to 30°C?

विकल्प

• 1.0°C/min

• 0.5°C/min

• 1.5°C/min

• 0.25°C/min

Answer 1

0.5°C/min

Explanation:

Applying Newton's Law:

C = \[\frac{(dT/dt)_1}{T_1-T_0}=\frac{2.0}{60-20}=\frac{2.0}{40}\] = 0.05 min^-1

At T₂ = 30°C:

\[\left(\frac{dT}{dt}\right)_2\] = C(T₂ ​− T₀) = 0.05 × (30 − 20) = 0.05 × 10 = 0.5°C/min

Since the temperature difference dropped from 40°C to 10°C (a factor of 4), the cooling rate also drops by a factor of 4, directly confirming Newton's Law.