Answer:
T=599.9°C
Explanation:
To solve this problem you can use the Newton's law of cooling
[tex]T=T_a+(T_0-T_a)e^{-kt}\\k=\frac{\alpha A}{m c_e}[/tex]
Ta: room temperature
T0: initial temperature
α: convective heat transfer coefficient
m: mass
c: specific heat
A: area
Hence, by replacing we have
[tex]V=(0.05m)^3=1.25*10^{-4}m^3\\m=V\rho=(1.25*10^{-4}m^3)(8530\frac{kg}{m^3})=1.06kg\\\\A=(0.5m)^2=0.25m^2\\\\k=\frac{(40*10^{-6}\frac{m^2}{s})(0.25m^2)}{(380\frac{J}{kgK})(1.06kg)}=2.48*10^{-8}s^{-1}\\\\T=(15\°C)+(600\°C-15\°C)e^{-(2.48*10^{-8}s^{-1})(180s)}=599.9\°C[/tex]
hope this helps!!
