Question

A hot plate of area 0.2 $m^2$ is maintained at a temperature of ${59}^{o}C$ by a 100 W heater when the room temperature is ${20}^{o}C$. The appropriate convection coefficient is 2.512 $(\Delta T{)}^{1/4}W/m^{2}K$. Taking heat transfer from plate is by convection and radiation, the percentage of heat lost by radiation is

• A. 27
• B. 51
• C. 56
• D. 49

Answer 1

The correct option is B 51

$A=0.2m^2,T_s={59}^{o}C,T_{\infty }={20}^{o}C$

Convection heat transfer coefficient

$\left(h\right)=2.512(\Delta T{)}^{1/4}$

$h=2.512(59-20{)}^{1/4}$

$=2.512\times (39{)}^{1/4}$

$h=6.2774W/m^2-K$

Convection heat transfer = hA $\Delta T$

$=6.2774\times 0.2\times (59-20)$

$=40.96W$

$\approx 49W$

Radiation heat transfer = total heat transfer - convection heat transfer
= 100 - 49 = 51W

% of heat loss by radiation

$=\frac{51}{100}\times 100=51$%