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KCET2012PhysicsThermal Properties of Matter

A hot body is allowed to cool. The surrounding temperature is constant at 30^ C . The takes time t₁ to cool from 70^ C to 68^ C and time t₂ to cool from 60^ C to 59.5^ C . Then

Options

  1. At ₂= t ₁
  2. Bt ₂=2 t ₁
  3. Ct ₂= 1 2 t ₁
  4. Dt₂=4 t₁

Correct answer

B. t ₂=2 t ₁

Step-by-step solution

By Newton's law of cooling - dT dt = ( T - T ₀ ) 2 t₁ = (90-30)=60 and 0.5 t₂ = (60-30)=30 From Eqs. (i) and (ii), t ₂=2 t ₁

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