Question

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if a liquid takes 30s in cooling from 95°c to 90°c and 70s in cooling from 55°c to 50°c then temp. of room is ?​

Answer 1

T₀ = room temperature,  T = temperature of the liquid at time t

dQ/dt = dT/dt = rate of cooling = - k [T - T₀]    --- (1)

dT/[T - T₀] = - k dt

  =>  Ln [(T₂ - T₀) / (T₁ - T₀)] = - k t

        (T₂ - T₀) = (T₁ - T₀) exp(- k t)

Given:   t = 30 sec, T₂ = 70°C   , T₁ = 80⁰C

       => Ln [(70 - T₀)/(80 - T₀)] = - k *30       --- (2)

 Also,    t = 70s, T₂ = 50⁰C,  T₁ = 60⁰C

       => Ln [(50 - T₀) /(60 - T₀)] = -k * 70      ---- (3)

This way it is difficult to solve the two equations (2) and (3). So we use an approximate method not involving Log and Exp functions.

(1) =>   dT/dt = - k (T - T₀)

     dT = 70 - 80 = -10°C,   dt = 30 sec,  T = avg temperature = 75°C

     So   - 10°C/30sec = - k (75°C - T₀)

     =>  3 k (75 - T₀) = 1          ---- (4)

Also,   dT = 50 -60 = -10°C,  dt = 70sec,  T = avg temperature = 55°C

     So  -10°C/70sec = - k (55 - T₀)

     =>  7 k (55 - T₀) = 1      ---- (5)

(4) / (5) =>    3 (75 - T₀) = 7 (55 - T₀)

                     4 T₀ = 385 - 225 = 160°C

                      T₀ = 40 ⁰C

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Answer 2

$\huge\mathcal{Hlo\:Reetu}$

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