if 4.35 kj of energy are needed to heat a sample of water from 62C into steam at 100.0C what is the mass of the sample
q1 - heat needed to go from 62 C to 100 C.
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial). Tf is 100 and Ti is 62.
q2 is heat needed to convert the sample of water at 100 to steam at 100.
q2 = mass H2O x heat vaporization.
So q1 + q2 = 4350 J
I show q1 above and q2 above. Set up the equation, mass H2O is the unknown you're looking for. You will need to look up specific heat H2O and heat vaporization if you don't already know it. Remember to keep the units straight; i.e., I've changer the 4.35 kJ to J. I suggest you keep everything in J. Post your work if you get stuck.
q2 =
To find the mass of the sample, we need to use the specific heat capacity formula:
Q = m * c * ΔT
where:
Q is the amount of energy (in kilojoules, kJ)
m is the mass of the sample (in grams)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the temperature change (in Celsius)
First, let's convert the given energy from kilojoules (kJ) to joules (J):
4.35 kJ * 1000 = 4350 J
Next, let's find the temperature change:
ΔT = final temperature - initial temperature
ΔT = 100.0°C - 62°C
ΔT = 38.0°C
Now, we can rearrange the formula to solve for mass (m):
m = Q / (c * ΔT)
m = 4350 J / (4.18 J/g°C * 38.0°C)
m ≈ 28.03 g
Therefore, the mass of the sample is approximately 28.03 grams.