If 150.0 g of Zinc at 100.0 C and 250.0g of liquid water at 10.0 C are mixed in an insulated container, what is the final temperature of the mixture? Czn= .39J/g C
Cwater= 4.184J/g C
Are we looking for the delta T?
If we are, how do you find Q (joules)
Thank you!
(Mzn*Czn*Tzn)+(Mwater*Cwater*Twater)=
Tfinal((Mzn*Czn)+(Mwater*Cwater))
To find the final temperature of the mixture, we need to use the principle of conservation of energy, which states that the energy gained by the zinc must equal the energy lost by the water.
First, let's calculate the heat gained by the zinc:
Qzinc = m × Czinc × ΔTzinc
Where:
m = mass of zinc = 150.0 g
Czinc = specific heat capacity of zinc = 0.39 J/g C
ΔTzinc = change in temperature of zinc = final temperature - initial temperature
Next, let's calculate the heat lost by the water:
Qwater = m × Cwater × ΔTwater
Where:
m = mass of water = 250.0 g
Cwater = specific heat capacity of water = 4.184 J/g C
ΔTwater = change in temperature of water = final temperature - initial temperature
Now, since the total energy gained by the zinc is equal to the total energy lost by the water, we can set the equations equal to each other:
Qzinc = Qwater
m × Czinc × ΔTzinc = m × Cwater × ΔTwater
Since we are looking for the final temperature, ΔT, we can rearrange the equation as follows:
(ΔTzinc) / (ΔTwater) = (Cwater × mwater) / (Czinc × mzinc)
Substituting the given values:
(ΔTzinc) / (ΔTwater) = (4.184 J/g C × 250.0 g) / (0.39 J/g C × 150.0 g)
Now we can solve for ΔTzinc / ΔTwater:
(ΔTzinc) / (ΔTwater) = 2781.0 / 58.5
= 47.60
Finally, we can calculate the final temperature:
final temperature = initial temperature + ΔTzinc
Substituting the given values:
final temperature = 100.0 C + 47.60 C
= 147.60 C
Therefore, the final temperature of the mixture is 147.60 degrees Celsius.