A 196-g aluminum calorimeter contains 530 g of water at 19.8° C. Aluminum shot with a mass equal to 273 g is heated to 100.0° C and is then placed in the calorimeter. Find the final temperature of the system, assuming that there is no heat transfer to the surroundings.
heat lost by Al shot + heat gained by H2O = 0
[mass Al x specific heat A x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute all of the numbers and solve for Tfinal.
what is Tinitial for which?
thanks i got it in the end
T final is the same for both.
Tinitial for Al is T it started; Tinitial for H2O is T it started.
To find the final temperature of the system, we can use the principle of conservation of energy, assuming that there is no heat transfer to the surroundings.
The total heat gained by the water and the calorimeter is equal to the total heat lost by the aluminum shot.
The heat gained by the water and the calorimeter is given by the equation: q1 = m1c1ΔT1, where m1 is the mass of water, c1 is the specific heat capacity of water, and ΔT1 is the change in temperature of water.
The heat lost by the aluminum shot is given by the equation: q2 = m2c2ΔT2, where m2 is the mass of aluminum, c2 is the specific heat capacity of aluminum, and ΔT2 is the change in temperature of the aluminum.
Since there is no heat transfer to the surroundings, we can set q1 equal to q2.
So we have the equation: m1c1ΔT1 = m2c2ΔT2
Plugging in the known values:
m1 = 530 g (mass of water)
c1 = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = final temperature - initial temperature of water = final temperature - 19.8°C
m2 = 273 g (mass of aluminum shot)
c2 = 0.897 J/g°C (specific heat capacity of aluminum)
ΔT2 = initial temperature of aluminum - final temperature
Now we can solve for the final temperature of the system.
m1c1ΔT1 = m2c2ΔT2
530 * 4.18 * (final temperature - 19.8) = 273 * 0.897 * (100 - final temperature)
Now we can simplify and solve this equation algebraically to find the final temperature.
First, distribute the multiplication:
2215.14 * (final temperature - 19.8) = 245.481 * (100 - final temperature)
Next, distribute and multiply:
2215.14 * final temperature - 43798.932 = 245.48100 - 245.481 * final temperature
Group the terms with the variable on one side:
2215.14 * final temperature + 245.481 * final temperature = 245.481 * 100 + 43798.932
Combine like terms:
2460.621 * final temperature = 24548.1 + 43798.932
2460.621 * final temperature = 68347.032
Divide both sides by 2460.621:
final temperature = 68347.032 / 2460.621
final temperature ≈ 27.8°C
Therefore, the final temperature of the system, assuming no heat transfer to the surroundings, is approximately 27.8°C.