how much energy is required to heat 1.00L of water from 20degrees celcius to 90degress celcius
Note the correct spelling of celsius.
q = mass x specific heat x (Tfinal-Tintial)
(Heat in Joules) = (mass in grams) x (specific heat) x (temperature change)
1 L = 1000 grams & specific heat of water = 4.184 J/gram Degree Celsius
Temp Change = 90 - 20 = 70 Deg C
Heat = 1000 g x 4.184 J/g Deg C X 20 Deg C =
83680 J
With 2 significant figures your answer would be 84 kJ
With 3 significant figures your answer would be 83.7 kJ
How many formul units of LiOH are in 134.5gof LiOH
To calculate the amount of energy required to heat 1.00L of water from 20 degrees Celsius to 90 degrees Celsius, you can use the formula:
Q = m * c * ∆T
Where:
Q = Energy (in joules)
m = Mass of the water (in grams)
c = Specific heat capacity of water (in joules per gram Celsius)
∆T = Change in temperature (in Celsius)
First, convert 1.00L of water to grams. The density of water is approximately 1g/mL, so 1L of water is equal to 1000 grams.
Next, identify the specific heat capacity of water. The specific heat capacity of water is about 4.18 joules per gram Celsius.
Finally, calculate the energy using the formula:
Q = (1000g) * (4.18 J/g°C) * (90°C - 20°C)
Q = 1000g * 4.18 J/g°C * 70°C
Q ≈ 293,200 Joules
Therefore, approximately 293,200 Joules of energy are required to heat 1.00L of water from 20 degrees Celsius to 90 degrees Celsius.