Calorimetry

Question :

The walls of a closed cubical box of edge 50cm are made of material of thickness 1mm and thermal conductivity 4X10^-4 cal s^-1cm^-1C^-1.

The inside temperature of the box is at 130 ⁰C and outside at 30 ⁰C . The temperature inside the interior of the box is maintained with the help of heater.

The heater is connected to 400 V DC voltage and has resistance R.

1) which of the following formula would be use to find heat loss/sec through walls of the cubical box

a) $\frac{\mathrm{KA}(T_2-T_1)}{d}$

b) $\frac{K(T_2-T_1){}_{}}{\mathrm{Ad}}$

c) $\frac{\mathrm{Kd}}{A}(T_2-T_1)$

d) None of these

2) Find the value of R

a)R=5.35 ohm

b)R=5 ohm

c) R=6 ohm

d) R=6.35 ohm

Solution:

The heat transmitted per second through the walls of the closed box is given by

$\frac{Q}{t}=\frac{\mathrm{KA}}{d}(T_2-T_1)$

Where A is the total area of the box as heat would be lost across all the sides

Now we have

K=4X10^-4 cal s^-1cm^-1C^-1

A=6X50X50 cm²

d=.1 cm

T₂-T₁=100

Substituting all the values in the above equation

Q/t= 6000 cal s^-1

The heat lost through the wall should be compensated through the heater to maintain the same temperature difference

Now heat produced by the Heater

$H=\frac{V^2}{R}\mathrm{Joule}=\frac{V^2}{4.2R}\mathrm{cal}$

$\frac{V^2}{4.2R}=6000$

We have V=400 V

or R=6.35 ohm