NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
A body takes 10 min to cool from 60 ℃ to 50 ℃ . If the temperature of the surroundings is 25 ℃ and 527 ℃ respectively. The final temperature of the body is
Options
- A48 ° C
- B46   ° C
- C49   ° C
- D42.85   ° C
Correct answer
D. 42.85   ° C
Step-by-step solution
According to Newton's law θ 1 - θ 2 t = K θ 1 + θ 2 2 - θ 0 ∴ 60 - 50 10 = K 60 + 50 2 - 25 ....(i) Let θ be the temperature after another 10 min ∴ 50 - θ 10 = K θ + 50 2 - 25 ....(ii) Dividing Eq.(i) by Eq. (ii), we get 10 50 - θ = 30 × 2 θ ∴ θ = 42.85 ℃