Answer:
Time = t = 30.01 minutes
Explanation:
The missing part of the question is that "The water heater power rating is 255 Watts, where 1 Watt = 1 J/s.
Firstly, calculating the work done by water heater to heat 5.5 L of water from 23° C to 43°C.
Work Done = Q = m C ΔT
where,
Q = Heat Transfer
m = mass of water
m = Density x Volume = 1 g/mL x 5.5 L = 1000g/L x 5.5L = 5500 g
ΔT = Temperature Difference = 43-23 = 20°C
C = Heat Capacity of Water = 4.186 J/g °C
Therefore,
[tex]Q = (5500g)(4.186J/gC) (20C)\\Q = 460460 J[/tex]
Since, Power can be given as Work done per unit time,
[tex]P = \frac{W}{t}[/tex]
or, in this case,
[tex]P = \frac{Q}{t}\\[/tex]
Here,
P = 255W or P = 255J (Since 1W = 1J)
[tex]P=\frac{Q}{t}\\255 = \frac{460460}{t}\\t = \frac{460460}{255}\\t = 1805.72 s\\t = \frac{1805.72}{60}\\t = 30.01 min[/tex]
