In an experiment, 200 g of aluminum at 100degrees celsius is mixed with 50.0g of water at 20.degrees celsius, with the mixture thermally isolated.(a)What is the equilibrium temperature?What are the entropy changes of(b)the aluminum,(c)the water,and(d)the aluminum-water system?
To calculate the equilibrium temperature and entropy changes in this experiment, we can use the principle of energy conservation, specific heat capacity, and the formula for entropy change.
Step 1: Calculate the heat gained or lost by each substance.
(a) Heat gained or lost by aluminum:
Q_aluminum = mass_aluminum * specific_heat_capacity_aluminum * change_in_temperature_aluminum
Given:
mass_aluminum = 200 g
specific_heat_capacity_aluminum = 0.897 J/g°C
change_in_temperature_aluminum = equilibrium_temperature - initial_temperature_aluminum
We need to find the equilibrium temperature, so let's assume it to be 'T' for now.
change_in_temperature_aluminum = T - 100°C
Q_aluminum = 200 g * 0.897 J/g°C * (T - 100°C)
(b) Heat gained or lost by water:
Q_water = mass_water * specific_heat_capacity_water * change_in_temperature_water
Given:
mass_water = 50.0 g
specific_heat_capacity_water = 4.18 J/g°C
change_in_temperature_water = equilibrium_temperature - initial_temperature_water
change_in_temperature_water = T - 20°C
Q_water = 50.0 g * 4.18 J/g°C * (T - 20°C)
Step 2: Apply the principle of energy conservation:
In a thermally isolated system, the total amount of heat gained is equal to the total amount of heat lost. So,
Q_aluminum + Q_water = 0
200 g * 0.897 J/g°C * (T - 100°C) + 50.0 g * 4.18 J/g°C * (T - 20°C) = 0
Step 3: Solve the equation for the equilibrium temperature (T).
Simplifying the equation above will give us the equilibrium temperature.
(179.4 T - 17940) + (209 T - 4180) = 0
388.4 T - 22120 = 0
388.4 T = 22120
T ≈ 56.99°C
Therefore, the equilibrium temperature is approximately 57°C.
Step 4: Calculate the entropy change of each substance.
(c) Entropy change of water:
ΔS_water = mass_water * specific_heat_capacity_water * ln(T_final / T_initial)
ΔS_water = 50.0 g * 4.18 J/g°C * ln(57°C / 20°C)
(d) Entropy change of aluminum:
ΔS_aluminum = mass_aluminum * specific_heat_capacity_aluminum * ln(T_final / T_initial)
ΔS_aluminum = 200 g * 0.897 J/g°C * ln(57°C / 100°C)
(e) Entropy change of the aluminum-water system:
ΔS_system = ΔS_aluminum + ΔS_water
Calculating these expressions will give you the entropy changes for each substance and the system.
To determine the equilibrium temperature and entropy changes of the aluminum, water, and aluminum-water system, we can use the principles of thermal equilibrium and the laws of thermodynamics. Let's break down each part of the question:
(a) Equilibrium Temperature:
To find the equilibrium temperature of the mixture, we need to consider the heat transfer between the aluminum and the water until they reach thermal equilibrium. We can use the principle of heat exchange:
q(aluminum) + q(water) = 0
The heat gained by the water (q(water)) is equal to the heat lost by the aluminum (q(aluminum)). The heat gained or lost is given by the formula:
q = m * c * ΔT
Where:
- q is the heat gained or lost
- m is the mass of the substance
- c is the specific heat capacity
- ΔT is the change in temperature
For aluminum:
m(aluminum) = 200 g
c(aluminum) = specific heat capacity of aluminum = 0.897 J/g°C
ΔT(aluminum) = unknown (equilibrium temperature - initial temperature = ΔT)
For water:
m(water) = 50.0 g
c(water) = specific heat capacity of water = 4.18 J/g°C
ΔT(water) = unknown (equilibrium temperature - initial temperature = ΔT)
Setting up the equation:
m(aluminum) * c(aluminum) * ΔT(aluminum) + m(water) * c(water) * ΔT(water) = 0
Solving for ΔT(aluminum) + ΔT(water) = 0:
200 g * 0.897 J/g°C * ΔT(aluminum) + 50.0 g * 4.18 J/g°C * ΔT(water) = 0
With the given masses and specific heat capacities, you can substitute the values and solve for the equilibrium temperature by rearranging the equation.
(b) Entropy Change of Aluminum:
To calculate the entropy change of aluminum (ΔS(aluminum)), we can use the equation:
ΔS(aluminum) = (q(aluminum) / T(aluminum))
Where:
- ΔS(aluminum) is the change in entropy of aluminum
- q(aluminum) is the heat gained or lost by the aluminum (calculated previously)
- T(aluminum) is the initial temperature of the aluminum
Substitute the known values to find the entropy change.
(c) Entropy Change of Water:
Similar to the entropy change of aluminum, you can use the formula:
ΔS(water) = (q(water) / T(water))
Where:
- ΔS(water) is the change in entropy of water
- q(water) is the heat gained or lost by the water (calculated previously)
- T(water) is the initial temperature of the water
Substitute the known values to calculate the entropy change.
(d) Entropy Change of Aluminum-Water System:
The entropy change of the aluminum-water system (ΔS(system)) is the sum of the entropy changes of aluminum and water:
ΔS(system) = ΔS(aluminum) + ΔS(water)
Add the previously calculated entropy changes to determine the total entropy change of the system.
Remember to use consistent units throughout the calculations.