Suppose 1400 J of heat energy are put into a sample of liquid water at 23°C, and the final temperature of the water is 78°C. How many grams of water are present?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
1400 J = mass x 4.184 x (78-23)
mass H2O = ?
To calculate the mass of water, we need to use the heat capacity formula:
Q = m * c * ΔT
Where:
Q = heat energy (in joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g·°C)
ΔT = change in temperature (final temperature - initial temperature)
We know the following values:
Q = 1400 J
c = 4.18 J/g·°C
ΔT = 78°C - 23°C = 55°C
Substituting these values into the formula, we can solve for the mass of water (m):
1400 J = m * 4.18 J/g·°C * 55°C
To solve for m, divide both sides of the equation by (4.18 J/g·°C * 55°C):
m = 1400 J / (4.18 J/g·°C * 55°C)
Calculating this gives us:
m ≈ 5.4 grams
Therefore, there are approximately 5.4 grams of water present in the sample.