BITSAT2011PhysicsThermal Properties of MatterActual
A copper sphere cools from 62^ C to 50^ C in 10 minutes and to 42^ C in the next 10 minutes. Calculate the temperature of the surroundings.
Options
- A18.01^ C
- B26^ C
- C10.6^ C
- D20^ C
Correct answer
B. 26^ C
Step-by-step solution
By Newton's law of cooling, ₁- ₂ t =-k [ ₁+ ₂ 2 - ₀ ] ....(1) A sphere cools from 62^ C to 50^ C in 10 ~min . 62-50 10 =- k [ 62+50 2 - ₀ ] ....(2) Now, sphere cools from 50^ C to 42^ C in next 10 ~min . 50-42 10 =- k [ 50+42 2 - ₀ ] ....(3) Dividing eq ^ n . (2) by (3) we get, 56- ₀ 46- ₀ = 1.2 0.8 or 0.4 ₀=10.4 Hence ₀=26^ C