Question

initial mass of water at its boiling point is 0.8kg. 4kw of heater is used to boil it completely. Assuming the specific latent heat of vaporization of water is 2Mj/kg. What is the time taken to vaporize all the water?​

Answer 1

Answer:

it'll take 400 sec

Explanation:

1kg water need 2MJ to completely vaporize

0.8kg water will need 1.6MJ ( by multiplying 0.8 and 2MJ)

now power = energy/time

time = energy / power

=1.6MJ / 4KW

= 400 sec

Answer 2

Given: Initial mass of water at its boiling point is 0.8kg. 4kw of heater is used to boil it completely. The specific latent heat of vaporization of water is 2 MJ/kg.

To find: The time taken to vaporize all the water.

Solution:

• If 1 kg of water requires 2 MJ of energy, then 0.8 hg will require,

        $\frac{0.8 * 2}{1} = 1.6 MJ$

• Power of the heater is the energy spent for unit time. It can be given by the formula,

        $P = \frac{E}{t}$

• Here, P is the power of the heater, E is the energy required and t is the time taken.

• The power of the heater is 4 kW, which is equal to 4000 W.
• The energy required is 1.6 MJ, which is 1600 J.

        $4000 W = \frac{1600 J}{t}$

         $t = 0.4 s$

Therefore, the time taken to vaporize all the water is 0.4 seconds.