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Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contains 1 L of water when full. Once boiling starts, it is observed that half of the water in the coffeemaker evaporates in 18 min. Determine the power rating of the electric heating element immersed in water. Also, determine how long it will take for this heater to raise the temperature of 1 L of cold water from 18°C to the boiling temperature. The enthalpy of vaporization of water at the saturation temperature of 100ºC iss hfg = 2256.4 kJ/kg. At an average temperature of (100 + 18)/2 = 59ºC, the specific heat of water is c = 4.18 kJ/kg·ºC and the density is about 1 kg/L.

Respuesta :

Answer:

[tex]P=1362\ W[/tex]

[tex]t'=251.659\ s[/tex] is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, [tex]T_i=18^{\circ}C[/tex]

time taken to vapourize half a liter of water, [tex]t=18\ min=1080\ s[/tex]

desity of water, [tex]\rho=1\ kg.L^{-1}[/tex]

So, the givne mass of water, [tex]m=1\ kg[/tex]

enthalpy of vaporization of water, [tex]h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}[/tex]

specific heat of water, [tex]c=4180\ J.kg^{-1}.K^{-1}[/tex]

Amount of heat required to raise the temperature of given water mass to 100°C:

[tex]Q_s=m.c.\Delta T[/tex]

[tex]Q_s=1\times 4180\times (100-18)[/tex]

[tex]Q_s=342760\ J[/tex]

Now the amount of heat required to vaporize 0.5 kg of water:

[tex]Q_v=m'\times h_{fg}[/tex]

where:

[tex]m'=0.5\ kg=[/tex] mass of water vaporized due to boiling

[tex]Q_v=0.5\times 2256.4[/tex]

[tex]Q_v=1.1282\times 10^{6}\ J[/tex]

Now the power rating of the boiler:

[tex]P=\frac{Q_s+Q_v}{t}[/tex]

[tex]P=\frac{342760+1128200}{1080}[/tex]

[tex]P=1362\ W[/tex]

Now the time required to heat to boiling point form initial temperature:

[tex]t'=\frac{Q_s}{P}[/tex]

[tex]t'=\frac{342760}{1362}[/tex]

[tex]t'=251.659\ s[/tex]

hamzaahmeds

This question involves the concepts of the specific heat capacity and latent heat of vaporization.

a. The power rating of the electric heater is "1362 W".

b. The time taken for the heater to raise the temperature of 1 L of water is "13.8 min".

a.

The power rating of the heater is given as follows:

[tex]P=\frac{Q_r+Q_v}{t}[/tex]

where,

P = Power rating = ?

t = time taken = 18 min = 1080 s

[tex]Q_r[/tex] = Heat required to raise the temperature of water = [tex]mC\Delta T[/tex]

m = mass of water = (Density)(Volume) = (1 kg/L)(1 L) = 1 kg

C = specific heat capacity of water = 4.18 KJ/kg.°C

ΔT = change in temperature = 100°C - 18°C = 82°C

[tex]Q_v[/tex] = heat required to vaporize half of the water = [tex]\frac{mL}{2}[/tex]

L = latent heat of vaporization of water = 2256.4 KJ/kg

Therefore,

[tex]P=\frac{mC\Delta T+\frac{mL}{2}}{t}\\\\P=\frac{(1\ kg)(4.18\ KJ/kg.^oC)(82^oC)+\frac{(1\ kg)(2256.4\ KJ/kg)}{2}}{1080\ s}\\\\P=\frac{342.76\ KJ+1128.2\ KJ}{1080\ s}[/tex]

P = 1.362 KW = 1362 W

b.

Now the time taken for the vaporization can be given as follows:

[tex]t_v = \frac{Q_v}{P}=\frac{1128.2\ KJ}{1.362\ KW}\\\\t_v=828.34\ s = 13.8\ min[/tex]

Learn more about specific heat capacity here:

https://brainly.com/question/2530523?referrer=searchResults

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