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Machine temperature after one hour of heating

Short answer

With initial rise 10 °C and final rise 80 °C, after 1 h the rise is θ = 80(1−e⁻⁰·⁵) + 10 e⁻⁰·⁵ ≈ 37.54 °C, so machine temperature ≈ 67.5 °C.

The initial temperature of a machine is 40°C. Calculate the temperature of the machine after 1 hour if its final steady temperature rise is 80°C and the heating time constant is 2 hours. The ambient temperature is 30°C.

When a machine starts with an initial rise θi\theta_i above ambient, its rise after time tt is

θ=θm(1−e−t/τ)+θi e−t/τ\theta = \theta_m\left(1-e^{-t/\tau}\right) + \theta_i\,e^{-t/\tau}

where θm\theta_m is the final steady rise and τ\tau the heating time constant. This follows from the heat balance: heat produced = heat stored + heat dissipated, P dt=Gh dθ+Sλθ dtP\,dt = G h\,d\theta + S\lambda\theta\,dt, whose solution is the expression above with τ=Gh/(Sλ)\tau = Gh/(S\lambda).

Data

  • Ambient temperature = 30 °C
  • Initial temperature = 40 °C, so initial rise θi=40−30=10 ∘\theta_i = 40 - 30 = 10\ ^\circC
  • Final steady rise θm=80 ∘\theta_m = 80\ ^\circC
  • τ=2\tau = 2 h, t=1t = 1 h

Calculation

e−t/τ=e−1/2=0.6065θ=80(1−0.6065)+10(0.6065)=31.48+6.07=37.54 ∘C\begin{aligned} e^{-t/\tau} &= e^{-1/2} = 0.6065\\ \theta &= 80(1-0.6065) + 10(0.6065)\\ &= 31.48 + 6.07 = 37.54\ ^\circ\text{C} \end{aligned}

Temperature of the machine =30+37.54=67.54 ∘= 30 + 37.54 = 67.54\ ^\circC.

 rise
 80 |- - - - - - - - - - - - - - -  final
    |              ___....-----
    |        _.--''
 37.5|- - -.'  (t = 1 h)
    |   /
 10 |_.'
    +-------------------------> t
    0      1 h

Answer: temperature after 1 hour ≈ 67.5 °C (rise of 37.5 °C above ambient).

✓ Answer checked by Claude Sonnet 5.5

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