Question

A 5.5-kW water heater operates at 240 V.

i. Should the heater circuit have a 20A or a 30A circuit breaker? [2 marks]

ii. Assuming 85% efficiency, how long will the heater take to heat the water in

a 0.2085-m3 tank from 20°C to 80°C? The density of water is 1,000 kg/m3

and the specific heat capacity is 4,200 J/kg/°C. ​

Answer 1

The time taken is
`t = 11.23 \ sec`

Step-by-step explanation:

From the question we are told

The power the water is
`P = 5.5KW = 5.5 *10^ {3} W`

The the voltage of the heater is is
`V = 240 V`

The volume of the heater
`Z = 0.2085\ m^3`

The specific heat of water is
`c_w = 4200 J /kg/^oC`

The initial temperature is
`T_1 = 20^oC`

The final temperature is
`T_2 = 80^oC`

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

The current of the heater is mathematically represented as

`I = (P)/(V )`

substituting values we have

`I = (5.5 *10^(3))/(240)`

`I = 22.91 \ A`

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

`Q = m c_w \Delta T`

Where m is the mass which is mathematically evaluated as

`m = \rho_w * Z`

`m =1000 * 0.2085`

`m =208.5 \ kg`

Therefore

`Q = 208.5 * 4.200 * (80 - 20)`

`Q = 52542 J`

Since the heater has an efficiency then the heat generate by the heater is

`Q_h = (52542)/(0.85)`

`Q_h =61814.1 J`

Now Power is mathematically represented as

`P = (Q)/(t)`

Where t is the time taken for heater to heat the water

=>
`t = (Q)/(P)`

Substituting values

`t = (61814.1)/(5.5*10^(3))`

`t = 11.23 \ sec`