Answer to Question #84789 in Physics for Rasheed sunday

Question

The heating coil rated 1kw is used to evaporated 700g of water at 100 degree.if specific latent heat of vaporization of water is 2.e×10^6, the time taken to evaporate all the water is?

Answer 1

Let's write the formula for the amount of heat required to change the state of water at

$$100 ℃$$

to a gas:

$$Q = mL,$$

here,

$$Q$$

is the amount of heat required to change the state of water at

$$100 ℃$$

to a gas,

$$m = 0.7 kg$$

is the mass of the water and

$$L = 2 \cdot 10^6 \frac{J}{kg}$$

is the specific latent heat of vaporization of water.

From the other hand, we can write the formula for the amount of heat that a coil provides at the rate of

$$1 kW$$

during time

$$t$$

:

$$Q = Pt,$$

here,

$$t$$

is the time taken to evaporate all the water.

Then, we can equate both expressions and find the time taken to evaporate all the water:

$$mL = Pt,$$ $$t = \frac{mL}{P} = \frac{0.7 kg \cdot 2 \cdot 10^6 \frac{J}{kg}}{1000 W} = 1400 s.$$

Answer:

$$t = 1400 s$$

.