Water is boiled at 1 atm pressure in a coffeemaker equipped with an immersion-type eletric heating element. The coffeemaker initially contains 1 kg of water. Once boiling has begun, it is observed that half of the water in the coffeemaker evaporates in 10 min. If the heat loss from the coffeemaker is negligible, the power rating of the heating element is
(a) 3.8 kW
(b) 2.2 kW
(c) 1.9 kW
(d) 1.6 kW
(e) 0.8 kW

Question

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Respuesta :

nafeesahmed nafeesahmed · Answer 1 · 2020-07-03T04:49:08+02:00

Given Information:

Pressure = 1 atm

Mass of water = 1 kg

time = 10 minutes

Required Information:

Power rating = ?

Answer:

The correct option is (c)

P = 1.9 kW

Explanation:

We want to find out the power rating of the heating element of the coffeemaker.

The power rating is given by

[tex]$ P = \frac{m \times h}{2\times t} $[/tex]

Where m is the mass of the water, h is the latent heat of vaporization of water, and t is the time in seconds.

From the standard table, the latent heat of vaporization of water at 1 atm and 100 °C is given by

h = 2257 kJ/kg

Time in seconds = 10×60 = 600 seconds

[tex]$ P = \frac{1 \times 2257}{2\times600 } $[/tex]

[tex]$ P = \frac{2257}{1200} $[/tex]

[tex]$ P = 1.88 \: kW$[/tex]

Rounding off yields

[tex]P = 1.9 \: kW[/tex]

Therefore, the correct option is (c) 1.9 kW