Answer:
The time taken is [tex]t = 11.23 \ sec[/tex]
Explanation:
From the question we are told
The power the water is [tex]P = 5.5KW = 5.5 *10^ {3} W[/tex]
The the voltage of the heater is is [tex]V = 240 V[/tex]
The volume of the heater [tex]Z = 0.2085\ m^3[/tex]
The specific heat of water is [tex]c_w = 4200 J /kg/^oC[/tex]
The initial temperature is [tex]T_1 = 20^oC[/tex]
The final temperature is [tex]T_2 = 80^oC[/tex]
The density of water is [tex]\rho_w = 1000 \ kg/m^3[/tex]
The current of the heater is mathematically represented as
[tex]I = \frac{P}{V }[/tex]
substituting values we have
[tex]I = \frac{5.5 *10^{3}}{240}[/tex]
[tex]I = 22.91 \ A[/tex]
So since the current produced is greater than 20 A hence the heater current rating would be 30A
The quantity of heat required to heat the water is mathematically represented as
[tex]Q = m c_w \Delta T[/tex]
Where m is the mass which is mathematically evaluated as
[tex]m = \rho_w * Z[/tex]
[tex]m =1000 * 0.2085[/tex]
[tex]m =208.5 \ kg[/tex]
Therefore
[tex]Q = 208.5 * 4.200 * (80 - 20)[/tex]
[tex]Q = 52542 J[/tex]
Since the heater has an efficiency then the heat generate by the heater is
[tex]Q_h = \frac{52542}{0.85}[/tex]
[tex]Q_h =61814.1 J[/tex]
Now Power is mathematically represented as
[tex]P = \frac{Q}{t}[/tex]
Where t is the time taken for heater to heat the water
=> [tex]t = \frac{Q}{P}[/tex]
Substituting values
[tex]t = \frac{61814.1}{5.5*10^{3}}[/tex]
[tex]t = 11.23 \ sec[/tex]
