Question 11.2: A block of metal is surrounded by adiabatic walls, except fo...
A block of metal is surrounded by adiabatic walls, except for a thermally conductive rod inside the wall connecting it to the environment. The initial temperature (at t = 0) of the metal block is T_{0} = 100.00°C, and its heat capacity is C = 600 kJ K^{-1}. The thermal conductivity of the 5 mm long rod having a cross section of 1 mm² is λ = 200 W m^{–1} K^{–1}. The environment has a constant temperature of T_{e} = 20.00°C. We follow the temperature difference of the environment and the metal block using a digital thermometer having a precision of 0.01°C.
Calculate the time needed for the metal block to cool down to 60°C and to cool down to the temperature of the surroundings as measured by the digital thermometer.
We can write (11.6) – Fourier’s law – for this problem as follows:
J_{z} (energy) = -λ \frac{dT}{dz}. (11.6)
\frac{1}{A} \frac{dQ}{dt} = -λ \frac{dT}{dz}.By supposing that the temperature gradient is uniform along the rod of length l, we can replace dT/dz by the finite difference ratio (T – T_{e})/l. The infinitesimal heat transfer dQ can be expressed with the heat capacity as dQ = CdT. Thus, the original differential equation can be written as follows:
\frac{1}{A} C \frac{\mathrm{d} T}{\mathrm{~d} t}=-\lambda \frac{T-T_{\mathrm{e}}}{l}By rearranging, we get the differential equation
\frac{\mathrm{d} T}{\mathrm{~d} t}=\lambda \frac{A}{C} \frac{T_{\mathrm{e}}-T}{l}
to solve. The solution of this equation with the boundary condition T_{0} = 100°C at t = 0 yields the function:
T(t)-T_{\mathrm{e}}=\left(T_{0}-T_{\mathrm{e}}\right) e^{-\frac{A \lambda t}{Cl}}
To get the time necessary to cool down to 60°C (by 40°C; half of the original difference), we have to substitute the difference T(t) – T_{e} = 40°C and solve the resulting exponential equation. The solution is 10 397.3 s, which is equivalent to 2 h and 58.3 min. The time necessary to cool down to the environmental temperature T_{e}= 20°C can be calculated by putting the temperature difference to T(t) – T_{e} ≤ 0.005°C, i.e., below the precision of the digital thermometer. The solution in this case is t ≥ 145205 s, which is equivalent to 2420 h and 2 min. The results indicate that the cool-down rate diminishes dramatically with the decrease of temperature difference.