1. Tardigrade
  2. Question
  3. Physics
  4. The walls of a closed cubical box of edge 5 cm are of a material of thermal conductivity 4× 104 cal s -1 cm-1 C-1 and thickness 1 mm. The interior is maintained at 100°C above the outside temperature by a heater placed inside the box and connected across a 400 V d.c. source. The resistance of the heater is

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Q. The walls of a closed cubical box of edge $5 \,cm$ are of a material of thermal conductivity $4\times 10^4 cal \, s^{ -1} cm^{-1} C^{-1}$ and thickness $1\, mm$. The interior is maintained at 100$^\circ$C above the outside temperature by a heater placed inside the box and connected across a $400\, V$ d.c. source. The resistance of the heater is

Current Electricity

Solution:

Using Q = $\frac{\lambda A(\theta_2 - \theta_1)t}{x} $ we get $\frac{Q}{t} = \frac{\lambda A ( \theta_2 - \theta_1)}{x}$
= $\frac{( 4 \times 10^{-4}) 6 \times 50 \times 50 \times 100}{0.1} $ ( $\because$ number of faces = 6 )
= 600 cal
Heat produced, H = $\frac{V^2}{R} \, i.e. \, 6000 \times 4.2 = \frac{400 \times 400}{R}$ or R = $\frac{400 \times 400}{4.2 \times 6000}$ = 6.35 $\Omega$