How much energy in kJ is needed to heat 7.0L of water from35 degrees Fahrenheit to boiling point 100 degrees Celsius. If the density of water is 1.0g/cm^3 and the heat capacity of water is Cp= 4.184J/g degrees Celsius
To calculate the energy needed to heat water, we need to follow several steps:
Step 1: Convert the initial temperature of water from Fahrenheit to Celsius.
To do this, we'll use the following formula:
Celsius = (Fahrenheit - 32) * (5/9)
Celsius = (35 - 32) * (5/9)
Celsius ≈ 1.7 degrees Celsius
Step 2: Calculate the mass of water.
We know that the density of water is 1.0 g/cm^3. Since the volume provided is in liters (L), we need to convert it to cubic centimeters (cm^3) using the conversion factor 1L = 1000cm^3.
Volume = 7.0L * 1000cm^3/L
Volume = 7000cm^3
Now we can calculate the mass using the density formula:
Mass = Volume * Density
Mass = 7000cm^3 * 1.0g/cm^3
Mass = 7000g
Step 3: Calculate the temperature change.
The temperature change is the difference between the final and initial temperatures:
ΔT = Final temperature - Initial temperature
ΔT = 100 degrees Celsius - 1.7 degrees Celsius
ΔT = 98.3 degrees Celsius
Step 4: Calculate the energy using the heat capacity formula:
Energy = Mass * ΔT * Cp
Energy = 7000g * 98.3 degrees Celsius * 4.184J/g degrees Celsius
Converting grams to kilograms, we get:
Energy = (7000g / 1000) kg * 98.3 degrees Celsius * 4.184J/g degrees Celsius
Calculating the energy:
Energy ≈ 289,632.04 J
Since 1 kJ is equal to 1000 J, we divide by 1000 to convert Joules to kilojoules:
Energy = 289,632.04 J / 1000
Therefore, the energy needed to heat 7.0L of water from 35 degrees Fahrenheit to 100 degrees Celsius is approximately 289.63 kJ.