Question

A cubical box of side 0.20 m is constructed from 0.012 m thick concrete panels of thermal

conductivity K=0.8W m−1 K−1

. A 100W heater is sealed inside the box and switched

on. The temperature of air outside the box is 20 ◦C. After a long period of time, the air

temperature inside the box will be close to​

Answer 1

according to formula,

$\quad\frac{dQ}{dt}=KA\frac{dT}{dx}$

where $\frac{dQ}{dt}$ is the rate of change of heat with respect to time, K is thermal conductivity , A is cross sectional area, dT is change in temperature and dx is the thickness by which heat flows.

here , A = 6(0.2m)² = 0.24m²

K = 0.8 W/m.K

dT = (T - 20°C) [ as heater is placed inside the cube of temperature will be higher inside the cube than outside]

dx = 0.012m

and dQ/dt = 100 W

now, 100 = 0.8 × 0.24 × (T - 20)/0.012

or, 100 × 0.012 = 0.192 × (T - 20)

or, 1200 = 192(T - 20)

or, 6.25 = T - 20

or, T = 26.25°C

hence, temperature inside the box will be 26.25°C

Answer 2

Explanation:

according to formula,

\quad\frac{dQ}{dt}=KA\frac{dT}{dx}

dt

dQ

=KA

dx

dT

where \frac{dQ}{dt}

dt

dQ

is the rate of change of heat with respect to time, K is thermal conductivity , A is cross sectional area, dT is change in temperature and dx is the thickness by which heat flows.

here , A = 6(0.2m)² = 0.24m²

K = 0.8 W/m.K

dT = (T - 20°C) [ as heater is placed inside the cube of temperature will be higher inside the cube than outside]

dx = 0.012m

and dQ/dt = 100 W

now, 100 = 0.8 × 0.24 × (T - 20)/0.012

or, 100 × 0.012 = 0.192 × (T - 20)

or, 1200 = 192(T - 20)

or, 6.25 = T - 20

or, T = 26.25°C

hence, temperature inside the box will be 26.25°C