10g of ice at 273K is added to 20 g of water at 90oC in an insulated flask. The heat of fusion of ice is 6 Kj/mol and the specific heat capacity of water is 4.2 J/K/g. ignoring the heat capacity of the flask;
Determine the final temperature of the system (3 marks)
Dtermine ∆S of the system (3 marks)
2. Calcul
what about ∆s of the system?
heat to melt ice + heat to move ice T to final T + heat to lower T of 90 C water to final T = 0
mols ice = 10/18 = 0.555
(0.555 mols ice x 6010 J/mol) + [10 x 4.2 J/mol*K(Tf-0)] +[20g x 4.2 J/mol*K x (Tf-90)] = 0
Solve for Tf = final temperature. Post your work if you get stuck.
Gives a simpler formula for calculating final temperature
Final temperature =306.5
Tf=306.5K
1. To determine the final temperature of the system, we need to consider the heat transfer that occurs between the ice and water.
First, let's calculate the heat transferred from the water to heat it up to the final temperature.
The formula to calculate heat transfer is q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the water:
q_water = 20 g * 4.2 J/g°C * (T_final - 90°C)
Next, let's calculate the heat transferred from the ice to melt it, using the heat of fusion.
The heat transferred from the ice is equal to the heat of fusion multiplied by the number of moles of ice. The formula to calculate the number of moles is moles = mass / molar mass.
The molar mass of water (H2O) is 18 g/mol, so the number of moles of ice is:
moles_ice = 10 g / 18 g/mol
Then, the heat transferred from the ice can be calculated as:
q_ice = 10 g * (6 kJ/mol / 18 g/mol)
Since the heat transfer is conserved, the heat transferred from the water to the ice is equal to the heat transferred from the ice to the water. Therefore:
q_water = q_ice
Now we can equate the equations for heat transfer:
20 g * 4.2 J/g°C * (T_final - 90°C) = 10 g * (6 kJ/mol / 18 g/mol)
Simplifying the equation, we get:
(T_final - 90°C) = 0.5 * (6 kJ/mol / 4.2 J/g°C)
Now, we can solve for T_final by rearranging the equation:
T_final = 90°C + (0.5 * (6 kJ/mol / 4.2 J/g°C))
Calculating the right-hand side of the equation gives us the final temperature.
2. To determine ΔS of the system, we need to consider the entropy change for both the ice and water.
For the ice, the entropy change (ΔS_ice) is given by the formula:
ΔS_ice = moles_ice * ΔS_fusion, where ΔS_fusion is the entropy change of fusion.
From the given data in the question, we know that the heat of fusion of ice is 6 kJ/mol. The entropy change of fusion can be calculated using the formula:
ΔS_fusion = ΔH_fusion / T_fusion, where ΔH_fusion is the heat of fusion and T_fusion is the melting point of ice.
Let's calculate ΔS_fusion first:
ΔS_fusion = (6 kJ/mol) / 273 K
Now, we can calculate ΔS_ice using the number of moles of ice we calculated earlier:
ΔS_ice = (10 g / 18 g/mol) * ((6 kJ/mol) / 273 K)
For the water, the entropy change (ΔS_water) can be calculated using the formula:
ΔS_water = m * c * ln(T_final / T_initial), where m is the mass, c is the specific heat capacity, T_final is the final temperature, and T_initial is the initial temperature.
Let's calculate ΔS_water using the given data:
ΔS_water = 20 g * 4.2 J/g°C * ln(T_final / 90°C)
Finally, to get the total entropy change (ΔS) of the system, we add the entropy changes of ice and water:
ΔS = ΔS_ice + ΔS_water
Calculating the equations gives us the entropy change of the system.