A portable electric water heater transfers 255 watts (W of power to 5.5 L of water, where 1 W=1J/s.How much time (in minutes) will it take for the water heater to heat the 5.5 L of water from 27∘C to 49∘C? (Assume that the water has a density of 1.0 g/mL.
To determine the time it takes to heat the water, we need to calculate the amount of heat required for the temperature change and then divide it by the power of the water heater.
1. Calculate the mass of water:
Density = 1.0 g/mL
Volume = 5.5 L = 5500 mL
Mass = Density * Volume = 5500 g
2. Calculate the heat required to raise the temperature:
Specific heat capacity of water = 4.18 J/g°C
Initial temperature = 27°C
Final temperature = 49°C
Heat required = (Mass) * (Specific heat capacity) * (Change in temperature)
= (5500 g) * (4.18 J/g°C) * (49°C - 27°C)
3. Convert heat required to joules to watts:
Since 1 W = 1 J/s, we need to convert the heat required from joules to watts.
Heat in joules = (Heat required)
Heat in watts = (Heat in joules) / (Time in seconds)
4. Calculate the time required in seconds:
Time in seconds = (Heat in joules) / (Power of the water heater in watts)
5. Convert the time from seconds to minutes:
Time in minutes = (Time in seconds) / 60
Now let's calculate each step:
1. Mass of water:
Mass = 5500 g
2. Heat required:
Heat required = (5500 g) * (4.18 J/g°C) * (49°C - 27°C)
3. Convert heat required to joules to watts:
Heat in joules = (Heat required)
Heat in watts = (Heat in joules) / (Time in seconds)
4. Calculate the time required in seconds:
Time in seconds = (Heat in joules) / (Power of the water heater in watts)
5. Convert the time from seconds to minutes:
Time in minutes = (Time in seconds) / 60
So, calculating these steps will give you the answer.
To solve this problem, we need to use the formula:
Q = mcΔT
Where:
Q is the heat energy transferred
m is the mass of water
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's calculate the mass of water. We know that the density of water is 1.0 g/mL. Since you have 5.5 L of water, we convert it to grams by multiplying it by the density:
Mass = Volume * Density
Mass = 5.5 L * 1000 g/L
Mass = 5500 g
Next, we need to calculate the change in temperature:
ΔT = Final Temperature - Initial Temperature
ΔT = 49°C - 27°C
ΔT = 22°C
Now, we can substitute the values into the formula and calculate the heat energy transferred:
Q = mcΔT
Q = 5500 g * 4.18 J/(g°C) * 22°C
Q = 10.922 kJ
Since power is the rate at which work is done, we can use the formula:
Power = Energy / Time
In this case, we are given the power (255 watts) and we want to find the time. Rearranging the formula:
Time = Energy / Power
Substituting the values:
Time = 10.922 kJ / 255 W
Since we want the answer in minutes, we need to convert the units appropriately:
Time = (10.922 kJ / 255 W) * (1 J/1000 kJ) * (1 min/60 s)
Time ≈ 0.04278 min
Therefore, it will take approximately 0.04278 minutes (or about 2.57 seconds) for the water heater to heat the 5.5 L of water from 27°C to 49°C.
q needed to heat the water is
q in Joules = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Then convert to 255 watts and q in Joules to time.