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
Answers
according to formula,
where 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
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