Question

The temperature difference between two sides of an iron plate, 5 cm thick, is 20 °C. Heat transmitted through the plate at the rate of 500 kcal per minute per square metre at steady state. Find the thermal conductivity of iron.​

Answer 1

The thermal conductivity of iron is 0.0014 kcal/m s k

Given,

$Q/At$ = 500kcal/min $m^{2}$ =$\frac{500}{60}$ kcal/s $m^{2}$

x = 5cm = 5×$10^{-2}$ m

($T_{1} -T_{2}$)= 20°C= 293K

To find: thermal conductivity of iron, k.

Formula: Q=$\frac{KA(T1-T2)}{x}$

Calculation: from the formula,

k= $\frac{Q}{At}$ × $\frac{x}{(T_{1} -T_{2}) }$

K= $\frac{500}{60}$ ×$\frac{5X10^{-2} }{293}$

K=0.0014 Kcal/m s k

Therefore, the thermal conductivity of iron is 0.0014 Kcal/m s k