Question

Water contained in a closed thin walled cylindrical copper tank, of radius 30cm and height 1m, is maintained at $60^\circ C$ by mean of an electric heater immersed in water, the outside temperature being $20^\circ C$. The tank’s outer curved surface is covered with 1cm thick felt $\left( {{K_{felt}} = 9 \times {{10}^{ - 5}}cal/s.cm.^\circ C} \right)$. Neglect all other losses. The wattage of the heater is:
1) 100
2) 1000
3) 220
4) 284

Answer 1

Hint: Here, the heat is transferred from one particle to another particle while the particle stays at its own position, this process is called conduction. Here, the water heater is heating up the water i.e. the water is carrying the heat energy. The change of heat with respect to time will be the wattage of the heater.

Complete step by step solution:
Here, we need to find out the formula for rate of change of heat (Wattage):
$\dfrac{{dQ}}{{dt}} = KA\dfrac{{dT}}{{dx}}$
Here:
A = Area;
K = Felt;
T = Temperature;
x = distance or length;
t = time;
Q = Heat;
The formula for the area is:
$Area = 2\pi r\left( {k + r} \right)$
Put in the given value:
$Area = 2 \times 3.14 \times 30 \times 130c{m^2}$ …(Here, K = 1m = 100cm )
$Area = 24492c{m^2}$
Put the above value in the relation$\dfrac{{dQ}}{{dt}} = KA\dfrac{{dT}}{{dx}}$
$\dfrac{{dQ}}{{dt}} = 9 \times {10^{ - 5}} \times 24492 \times \dfrac{{\left( {60 - 20} \right)}}{1}$
Do, the necessary calculation
$ \Rightarrow \dfrac{{dQ}}{{dt}} = 9 \times {10^{ - 5}} \times 24492 \times \dfrac{{40}}{1}$
The wattage comes out to be:
$ \Rightarrow \dfrac{{dQ}}{{dt}} = 88.17$
$ \Rightarrow \dfrac{{dQ}}{{dt}} \approx 100$

Therefore, Option “1” is correct. The wattage of the heater is 100.

Note: Here, we need to apply the formula for conduction and solve for the unknown. The height of the copper tank would be the height of the felt covering the tank and the height of the copper vessel. Put in the given values and solve for the unknown.