MHT CET202520 Apr 2025Evening ShiftPhysicsThermal Properties of MatterActual
A metal sphere cools at a rate of 1.5^ C / min when its temperature is 80^ C . When the temperature of the sphere is 40^ C , its rate of cooling is 0.3^ C / min . The temperature of the surrounding ( ₀ ) is
Options
- A30^ C
- B35^ C
- C25^ C
- D27^ C
Correct answer
A. 30^ C
Step-by-step solution
According to Newton's Law of Cooling, the rate of cooling is proportional to the temperature difference: d dt = -K( - ₀) Given two measurements: At ₁ = 80^ C, cooling rate is 1.5^ C/min At ₂ = 40^ C, cooling rate is 0.3^ C/min Dividing the corresponding equations: 1.5 0.3 = 80 - ₀ 40 - ₀ Solving for ₀ : 5(40 - ₀) = 80 - ₀ 200 - 5 ₀ = 80 - ₀ 120 = 4 ₀ ₀ = 30^ C The surrounding temperature is therefore 30^ C. Final answer: A