A 300.0g sample of ice at -30.OC is mixed with 300.0G of water at 90C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capazities of H2O(s) and H20(L) are 2.03 and 4.18 j/g*C respectivly, and the entahply fusion for ice is 6.02 kJ/mol
heat gained by ice from -30 to 0C + heat gained by melting ice + heat gained by water from ice @ 0 C moving to higher T + heat lost by H2O @ 90C = 0
[mass ice x specific heat ice x (Tfinal-Tinitial)] + [mass ice x delta Hfusion] + [mass melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass warm water x specific heat H2O x (Tfinal-Tinitial)]=0
You have values for everything there except Tf, solve for that.
133.317 cal/grams
To calculate the final temperature of the mixture, we need to consider the heat gained or lost by both the ice and water.
Step 1: Calculate the heat gained by the ice during the phase change from solid to liquid.
The enthalpy of fusion for ice (ΔHfusion) is given as 6.02 kJ/mol. We need to convert the mass of the ice to moles.
The molar mass of water is 18.015 g/mol, so the number of moles of ice is:
moles = mass / molar mass
moles = 300.0 g / 18.015 g/mol
moles ≈ 16.64 mol
Therefore, the heat gained by the ice during the phase change can be calculated as:
Q1 = moles * ΔHfusion
Q1 = 16.64 mol * 6.02 kJ/mol
Q1 ≈ 100.17 kJ
Step 2: Calculate the heat gained or lost by the water during the temperature change.
The specific heat capacity of water (Cwater) is given as 4.18 J/g°C. We can calculate the heat gained or lost by the water using the formula:
Q2 = mass * specific heat capacity * temperature change
Temperature change = final temperature - initial temperature
Initial temperature of the water (T1) = 90°C
Final temperature of the mixture (Tfinal) = ?
Mass of the water (mwater) = 300.0 g
Q2 = mwater * Cwater * (Tfinal - T1)
Q2 = 300.0 g * 4.18 J/g°C * (Tfinal - 90°C)
Step 3: Calculate the final temperature of the mixture.
Since there is no heat loss to the surroundings, the heat gained by the ice (Q1) must be equal to the heat gained by the water (Q2).
Q1 = Q2
100.17 kJ = 300.0 g * 4.18 J/g°C * (Tfinal - 90°C)
Now solve for Tfinal:
Tfinal - 90°C = (100.17 kJ) / (300.0 g * 4.18 J/g°C)
Tfinal - 90°C ≈ 79.5°C
Tfinal ≈ 79.5°C + 90°C
Tfinal ≈ 169.5°C
Therefore, the final temperature of the mixture, assuming no heat loss to the surroundings, is approximately 169.5°C.
To calculate the final temperature of the mixture, we can use the principle of conservation of energy.
First, let's calculate the energy required to bring the ice sample at -30°C to its melting point of 0°C:
Energy = mass x specific heat capacity x change in temperature
Energy = 300.0 g x 2.03 J/g°C x (0°C - (-30°C))
Energy = 300.0 g x 2.03 J/g°C x 30°C
Energy = 18,270 J
Next, let's calculate the energy released when the ice melts at 0°C:
Energy = moles x enthalpy fusion
First, we need to determine the number of moles of ice:
Molar mass of H2O = 18.015 g/mol
Number of moles = mass / molar mass = 300.0 g / 18.015 g/mol = 16.649 mol
Energy = 16.649 mol x 6.02 kJ/mol x 1000 J/kJ
Energy = 100,065.98 J
Now, let's calculate the energy required to heat the resulting water from 0°C to the final temperature:
Energy = mass x specific heat capacity x change in temperature
Energy = 300.0 g x 4.18 J/g°C x (final temperature - 0°C)
Since there is no heat loss to the surroundings, the energy released by the ice melting must equal the energy gained by the resulting water:
18,270 J + 100,065.98 J = 300.0 g x 4.18 J/g°C x (final temperature - 0°C)
Simplifying the equation:
118,335.98 J = 1254 J/g°C x (final temperature)
Dividing both sides by 1254 J/g°C:
(final temperature) = 94.43°C
Therefore, the final temperature of the mixture is approximately 94.43°C.