Answer:
the time required for the cooling process = 0.1635 hours
Step-by-step explanation:
We are given;
Thermal conductivity;k = 40 W/m·K
Density;ρ = 7800 kg/m³
Specific heat capacity;c = 600 J/kg·K
h = 25 W/m².k
Diameter;D = 10mm = 0.01m
T_(∞) = 325K
T_i = 1150K
T = 450K
The formula for time required for the cooling process is given by;
t = (ρVc/hA)[In((T_i - T_(∞))/(T - T_(∞)))]
Where;
V is volume = πD³/6
A is Area = πD²
Other terms remain are previously stated.
Thus;
t = (ρ(πD³/6)c/(h(πD²))[In((T_i - T_(∞))/(T - T_(∞)))]
This is simplified into;
t = (ρDc/(6h)[In((T_i - T_(∞))/(T - T_(∞)))]
Plugging in the relevant values to obtain;
t = (7800*0.01*600/(6*25)[In((1150 - 325)/(450 - 325))]
t = 312In(825/125)
t = 588.766 seconds = 0.1635 hours