Question

The heating element in an electric kettle is rated as 2.0 kW. If the water in the kettle is at 100.0 °C, what volume of water will be converted into steam in one minute? The specific latent heat of vaporization of the water is 2,257,000 J/kg and the

3 density of water is 1,000 kg/m .

Answer 1

The volume is
`V =5.32 *10^(-5) \ m^3`

Step-by-step explanation:

From the question we are told that

The power of the heating element is
`P = 2.0 kW = 2.0 *10^3 \ W`

The temperature of the water in the kettle is
`T _w = 100^oC`

The time to convert water to steam is t = 1 minute = 60 sec

The specific latent heat of vaporization is
`H_v = \ 2,257,000 J/kg`

The density of water is
`\rho_w = 1000\ kg/m^3`

The power of the heating element is mathematically represented as

`P = (E)/(t)`

Where E Energy generated by the heating element in term of heat

`E = Pt`

substituting values

`E = 2.0 *10^(3) * 60`

`E = 120000 J`

Now

The latent heat of vaporization is mathematically represented as

`H_v = (E)/(m)`

Where m is the mass of water converted to steam

So

`m = (E)/(H_v)`

substituting values

`m = (120000)/(2257000)`

`m = 0.0532\ kg`

The volume of water converted to steam is mathematically evaluated as

`V = (m )/(\rho_w)`

substituting values

`V = (0.0532)/(1000)`

`V =5.32 *10^(-5) \ m^3`