Newton’s law of cooling leads us to the following expression,

Physics · JEE Main · NTA Exams — Thermodynamics

1. (θ – θ₀) = Kt + C
2. log (θ – θ₀) = –Kt + C
3. log θ = Kt + C
4. θ = Kθ₀ + C

Answer

(B) log ( θ – θ 0 ) = –Kt + C

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