To solve this question, we can use the principle of conservation of energy, specifically the heat equation:
Q = mcΔT
Q represents the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we have three components: copper, water, and the aluminum calorimeter. We'll calculate the heat exchanged for each component and then use the equation to find the mass of copper.
Let's write the equation for each component:
For the copper: Qcopper = mcoppercopperΔTcopper
For the water: Qwater = mwaterwaterΔTwater
For the aluminum calorimeter: Qcalorimeter = mcaluminumcalorimeterΔTcalorimeter
Given the temperatures:
Tcopper = 90°C
Twater = 15°C
Tfinal = 25°C
Given the masses:
mwater = 200g
mcalorimeter (aluminum) = 100g
Specific heat capacity values:
ccopper = 0.39 J/g°C (copper specific heat capacity)
cwater = 4.18 J/g°C (water specific heat capacity)
ccalorimeter (aluminum) = 0.897 J/g°C (aluminum specific heat capacity)
We can substitute these values into the equations and then equate the total heat transferred to zero, since it is a closed system:
Qcopper + Qwater + Qcalorimeter = 0
Let's solve the equations step by step:
For the copper:
Qcopper = mcopperccopperΔTcopper
= mcopper(0.39)(25 - 90)
= -54.45mcopper (Note: the negative sign appears because the copper loses heat)
For the water:
Qwater = mwatercwaterΔTwater
= (200)(4.18)(25 - 15)
= 4180
For the aluminum calorimeter:
Qcalorimeter = mcaluminumccalorimeterΔTcalorimeter
= (100)(0.897)(25 - 15)
= 897
Now, let's substitute these values into the equation and solve for the mass of copper:
Qcopper + Qwater + Qcalorimeter = 0
-54.45mcopper + 4180 + 897 = 0
Combine like terms:
-54.45mcopper + 5077 = 0
-54.45mcopper = -5077
Divide by -54.45:
mcopper = -5077 / -54.45
mcopper ≈ 93.18g
Therefore, adding approximately 93.18g of copper at 90°C to 200g of water at 15°C contained in a 100g aluminum calorimeter will result in a final temperature of 25°C.