A piece of lead having a mass of 200.0 g and a temperature of 100.0 ˚C is placed in a cup of water initially at 22 ˚C. After stirring the final temperature of the water and lead is 25 ˚C. Calculate the heat energy that was lost by the lead to the water. Calculate the amount of water in the cup. C Pb = 0.160 J/g C
heat lost by Pb.
q = mass Pb x specific heat Pb x (Tfinal-Tinital) = ?
part 2.
mass Pb x specific heat Pb x (Tfinal-Tinital) + mass H2O x specific heat H2O x (Tfinal - Tinitial)
Substitute into that equation and solve for mass H2O
Post your work if you get stuck.
To calculate the heat energy lost by the lead to the water, we can use the equation:
Q = m x C x ΔT
Where:
Q is the heat energy (in joules)
m is the mass of the substance (in grams)
C is the specific heat capacity of the substance (in J/g ˚C)
ΔT is the change in temperature (in ˚C)
For the lead, we have:
m = 200.0 g (mass of lead)
C Pb = 0.160 J/g ˚C (specific heat capacity of lead)
ΔT = 100.0 ˚C - 25.0 ˚C (change in temperature)
Calculating the Q for the lead:
Q = 200.0 g x 0.160 J/g ˚C x (100.0 ˚C - 25.0 ˚C)
Q = 200.0 g x 0.160 J/g ˚C x 75.0 ˚C
Q = 2400 J
Therefore, the heat energy lost by the lead to the water is 2400 joules.
Now, to calculate the amount of water in the cup, we can use the equation:
Q = m x C x ΔT
Rearranging this equation to solve for mass, we have:
m = Q / (C x ΔT)
For water, we have:
Q = -2400 J (negative because it gained the energy lost by the lead)
C H2O = 4.18 J/g ˚C (specific heat capacity of water)
ΔT = 25.0 ˚C - 22.0 ˚C (change in temperature)
Substituting the values into the equation:
m = -2400 J / (4.18 J/g ˚C x 3.0 ˚C)
m ≈ -2400 J / 12.54 J/g
m ≈ -191.22 g
Since mass cannot be negative, we can take the absolute value to get the amount of water in the cup:
m ≈ 191.22 g
Therefore, the amount of water in the cup is approximately 191.22 grams.