Question

If a liquid takes 30 sec in cooling from 80°C to 70°C and 70sec in cooling from 60°C to 50°C then find the room temperature

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

Answer 2

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