JEE Main2014PhysicsThermal Properties of MatterActual
Hot water cools from 60^ C to 50^ C in the first 10 minutes and to 42^ C in the next 10 minutes. The temperature of the surroundings is:
Options
- A25^ C
- B10^ C
- C15^ C
- D20^ C
Correct answer
B. 10^ C
Step-by-step solution
By Newton's law of cooling ₁- ₂ t =- K [ ₁+ ₂ 2 - ₀ ] where ₀ is the temperature of surrounding. Now, hot water cools from 60^ C to 50^ C in 10 minutes, 60-50 10 =-K [ 60+50 2 - ₀ ] Again, it cools from 50^ C to 42^ C in next 10 minutes. 50-42 10 =- K [ 50+42 2 - ₀ ] Dividing equations (i) by (ii) we get aligned & 1 0.8 = 55- ₀ 46- ₀ & 10 8 = 55- ₀ 46- ₀ &460-10 ₀=440-8 ₀ &2 ₀=20 & ₀=10^ c aligned