If a water heater has a capacity of 54.4 gallons and a 4910 W heating element, how much time is required to raise the temperature of the water from 19.8oC to 60.0oC? Assume that the heater is well insulated and no water is withdrawn from the tank during this time.
To find the time required to raise the temperature of the water, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy required (in joules)
m is the mass of water (in kilograms)
c is the specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
1. First, let's calculate the mass of water in the tank. To do this, we'll use the formula:
mass = volume × density
The volume of water is given as 54.4 gallons. To convert this to liters, we multiply by 3.78541 (since 1 gallon is approximately equal to 3.78541 liters). Finally, to convert liters to kilograms, we need to multiply by the density of water, which is 1 kg/L.
So, the mass of water is:
mass = 54.4 gallons × 3.78541 L/gallon × 1 kg/L
mass ≈ 205.663 kg
2. Next, let's calculate the change in temperature:
ΔT = 60.0oC - 19.8oC
ΔT = 40.2oC
3. The specific heat capacity of water is 4.186 J/g·°C. To convert this to joules per kilogram per degree Celsius, we need to divide by 1000 since there are 1000 grams in a kilogram. So, the specific heat capacity of water is approximately 4.186 J/kg·°C.
4. Now we can substitute the values into the formula and solve for the heat energy required (Q):
Q = mcΔT
Q = (205.663 kg) × (4.186 J/kg·°C) × (40.2oC)
Q ≈ 341683.0276 J
5. Lastly, let's calculate the time required to raise the temperature using the formula:
time = Q / power
The power of the heating element is given as 4910 W.
time = (341683.0276 J) / (4910 W)
time ≈ 69.61 seconds
Therefore, it would take approximately 69.61 seconds to raise the temperature of the water from 19.8oC to 60.0oC.