How much heat is required to raise the temperature
of exactly one liter of water fro m 60 °F to 180 °F?
answer: 279 kj/mol
Convert 60 F to C. Convert 180 F to C.
q = mass x specific heat x (Tfinal-Tinitial).
Tf and Ti to be in celsius.
mass of 1L H2O = 1000 g.
The answer is not right. The answer is 279 kJ (for 55.5 mol) and not 279 kJ/mol.
I just copied the units wrong from the practice test we have.
To calculate the amount of heat required to raise the temperature of water, we need to use the specific heat capacity of water. The specific heat capacity of water is approximately 4.184 joules per gram per degree Celsius (J/g°C) or 1 calorie per gram per degree Celsius (cal/g°C).
First, let's convert the given temperatures from Fahrenheit to Celsius. The formula to convert Fahrenheit to Celsius is:
°C = (°F - 32) x (5/9)
For the initial temperature of 60 °F:
°C = (60 - 32) x (5/9) = 27.78 °C
For the final temperature of 180 °F:
°C = (180 - 32) x (5/9) = 82.22 °C
Now, we can calculate the change in temperature:
ΔT = Final temperature - Initial temperature = 82.22 °C - 27.78 °C = 54.44 °C
Since we want to find the heat required for exactly one liter of water (1000 grams), we multiply the specific heat capacity of water by the mass and the change in temperature:
Heat = mass x specific heat capacity x ΔT
Heat = 1000 g x 4.184 J/g°C x 54.44 °C
Now, let's convert joules to kilojoules (kJ). Since 1 kilojoule = 1000 joules, we divide the result by 1000:
Heat = (1000 g x 4.184 J/g°C x 54.44 °C) / 1000
Heat = 227509.92 J / 1000 = 227.51 kJ
So, approximately 227.51 kJ of heat is required to raise the temperature of one liter of water from 60 °F to 180 °F.