Question

A portable electric water heater transfers 255 watts (W) of power to 5.5 L of water, where 1W=1J/s. How much time (in minutes) will it take for the water heater to heat the 5.5L of water from 23 degrees Celsius to 43 degrees Celsius? (Assume that the water has a density of 1.0g/mL)

Answer 1

Q = mcΔT

where:
Q = heat energy (Joules)
m = mass of water (grams)
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature (°C)

The mass of 5.5 L of water is 5.5 L x 1 g/mL = 5500 g

So, the heat energy required to raise the temperature of the water is:
Q = (5500 g)(4.184 J/g°C)(43°C - 23°C)
Q = (5500)(4.184)(20)
Q = 4,638,400 J

Now we can divide this value by the power of the electric water heater to get the time it will take to heat the water:
time = Q / power
time = 4,638,400 J / 255 W
time = 18,211 s
time = 18,211 s / 60 s/min = 303.52 min

So it will take about 303.52 minutes for the electric water heater to heat the 5.5 L of water from 23 degrees Celsius to 43 degrees Celsius.