Chapter 18, Problem 70GP

To determine

The minimum wattage that the heater must have.

Answer to Problem 70GP

  $1463W$

Explanation of Solution

Given:

Specific heat, $c=0.17kcal/kg\cdot °C$

Temperature change, $\Delta T=20°C-5°C=15°C$

Volume is $62m^3$

Formula used:

Heat energy is $\Delta E=Q=mc\Delta T$ .

Where, $c$ is a specific heat, $\Delta T$ is the temperature change and $m$ is the mass of the object.

Calculation:

From the equation of heat energy: $\Delta E=Q=mc\Delta T$

Therefore, $\frac{Q}{t}=\frac{m}{t}c\Delta T\mathrm{...}(1)$

It is known that volume of the room is $62m^3$ and air entered in the room twice of $62\frac{m^3}{h}$

So, volume is $2\times 62\frac{m^3}{h}=124\frac{m^3}{h}$

At normal temperature pressure density of the air $1.29\frac{kg}{m^3}$

From the equation $(1)$ , calculate the heat energy:

  $Q=\left(124\frac{m^3}{h}\right)\left(1.29\frac{kg}{m^3}\right)\left(0.17\frac{kcal}{kg\cdot C°}\right)\left(15C°\right)$

  $Q=408\frac{kcal}{h}$

Now, by using the values of heat energy calculate the required power:

Required power $=408\frac{kcal}{h}+850\frac{kcal}{h}$

Required power $=1463W$

Conclusion: The required power is $1463W$ .