A 200g block of copper at 90o
C is dropped into 400g of water at 27o
C contained
in a 300g glass beaker at 27o
C. What is the final equilibrium temperature of the
mixture?
cCu = 0.0924 cal/g°C
cglass = 0.2 cal/g°C
cwater = 1 cal/g°C
The sum of heats gained is zero.
heat gained by copper+heatgainedwater+hatgainedglass=0
mCu*Ccu(Tf-90)+mglass*Cglas*(Tf-27)+mwater*cwat(Tf-27)=0
solve for Tf
To find the final equilibrium temperature of the mixture, we can use the principle of conservation of energy.
Let's calculate the heat gained or lost by each component separately:
1. Heat gained or lost by copper:
Q_cu = m_cu * c_cu * ΔT_cu
Q_cu = 200g * 0.0924 cal/g°C * (T_eq - 90°C)
2. Heat gained or lost by water:
Q_water = m_water * c_water * ΔT_water
Q_water = 400g * 1 cal/g°C * (T_eq - 27°C)
3. Heat gained or lost by the glass beaker:
Q_glass = m_glass * c_glass * ΔT_glass
Q_glass = 300g * 0.2 cal/g°C * (T_eq - 27°C)
According to the principle of conservation of energy, the total heat gained by one component equals the total heat lost by another component, assuming no heat is gained or lost to the surroundings.
Q_cu + Q_glass + Q_water = 0
200g * 0.0924 cal/g°C * (T_eq - 90°C) + 300g * 0.2 cal/g°C * (T_eq - 27°C) + 400g * 1 cal/g°C * (T_eq - 27°C) = 0
Now we can solve this equation to find the equilibrium temperature (T_eq).
To find the final equilibrium temperature of the mixture, we can apply the principle of energy conservation.
The total heat gained by the system must be equal to the total heat lost by the system. The formula for calculating heat gained or lost is:
Q = mcΔT
Where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Let's calculate the heat lost by the copper block, the heat lost by the glass beaker, and the heat gained by the water.
Heat lost by the copper block:
Qcopper = mcopper * ccopper * (Tfinal - Tinitial)
Qcopper = 200g * 0.0924 cal/g°C * (Tfinal - 90°C)
Heat lost by the glass beaker:
Qglass = mglass * cglass * (Tfinal - Tinitial)
Qglass = 300g * 0.2 cal/g°C * (Tfinal - 27°C)
Heat gained by the water:
Qwater = mwater * cwater * (Tfinal - Tinitial)
Qwater = 400g * 1 cal/g°C * (Tfinal - 27°C)
Since the total heat gained must equal the total heat lost, we can set up the equation:
Qcopper + Qglass = Qwater
Substituting the respective formulas:
200g * 0.0924 cal/g°C * (Tfinal - 90°C) + 300g * 0.2 cal/g°C * (Tfinal - 27°C) = 400g * 1 cal/g°C * (Tfinal - 27°C)
Simplifying the equation:
18.48Tfinal - 16.5720 + 60Tfinal - 4860 = 400Tfinal - 10800
Combining like terms:
78.48Tfinal - 4876.5720 = 400Tfinal - 10800
Moving the terms to one side:
-321.52Tfinal = -5923.428
Dividing both sides by -321.52:
Tfinal ≈ 18.4109°C
Therefore, the final equilibrium temperature of the mixture is approximately 18.4109°C.