Answer:
788°C
Step-by-step explanation:
Metal palte details
Thickness = t=10 mm =0.01 m.
Thermal conductivity= k =200 W/mC
Heat capacity C=900 J/kg-k.
Initial temp Ti =22 C.
Time for which it has been kept in oven =t=2 min =120 sec.
Air properties.
Temp of air in oven =Ta =900 C
Heat transfer coefficient =h =190 W/m-k.
Temp of block after 2 min= T =?
In problem, it is not given block dimension. Let us assume that Block is 1 m in legth and 1 m in width.
Volume of block =V= 1*1*0.01 =0.01 m3 .(We know thckness =0.01 m).
Surface area of block = A=2(1*1+1*0.01+0.01*1)=2.04 m2.
Biot number for this problem is 0.004656 which is less than 0.1. So Lumped capacitance method is applicable to this problem, according to formulae,
T - Tair /Ti - Tair =e(-hAt/density* C* V) .
T - 900/(22-900) = e(-190*2.04*120 /2500*900*0.01) .
T - 900/(22-900) =e-2.0672 .
T -900 = - 110.09
Temp of block after 2 min =788 °C.