How much ice (in grams) be added to lower the temperature of 355 mL of water from 25 ∘C to 5 ∘C ? (Assume the following: the density of water is 1.0 g/mL, the ice is at 0 ∘C, the heat of fusion of ice is 6.02 kJ/mol, and the volume of water produced by the melting ice is negligible.)
First, let's determine how much heat energy (Q) will be required to lower the temperature of water from 25°C to 5°C.
The formula to calculate heat energy is:
Q = m * c * ΔT
Where m is the mass of water, c is the specific heat capacity of water, which is 4.18 J/g°C, and ΔT is the change in temperature.
First, we need to convert the volume of water (355 mL) to mass. Since the density of water is 1.0 g/mL, the mass of water is:
m = 355 mL * 1.0 g/mL = 355 g
Now, let's find the change in temperature:
ΔT = T_final - T_initial = 5°C - 25°C = -20°C
Now we can calculate the heat energy:
Q = m * c * ΔT = 355 g * 4.18 J/g°C * (-20°C) = -29666 J
Now, we need to determine how many grams of ice will be required to absorb this heat energy.
We know the heat of fusion of ice is 6.02 kJ/mol. First, let's convert this to J/g. The molar mass of water is 18.015 g/mol, so:
Heat of fusion = (6.02 kJ/mol) * (1000 J/kJ) / (18.015 g/mol) = 334.19 J/g
Now we can calculate the mass of ice:
m_ice = Q / Heat of fusion = -29666 J / 334.19 J/g ≈ 88.8 g
So, you would need to add approximately 88.8 grams of ice to lower the temperature of 355 mL of water from 25°C to 5°C.
To determine the amount of ice needed to lower the temperature of water, we can use the formula for heat transfer:
q = m * C * ΔT
Where:
q is the heat transferred,
m is the mass of the substance (water or ice),
C is the specific heat capacity of the substance,
ΔT is the change in temperature.
First, let's calculate the heat transferred when the temperature of the water is lowered from 25 ∘C to 0 ∘C.
q1 = m1 * C1 * ΔT1
Since the water is going through a phase change from liquid to solid, we need to account for the heat of fusion (Hf) of ice. The heat transferred during this phase change is given by:
q2 = m2 * Hf
Finally, let's calculate the heat transferred when the temperature of the ice is lowered from 0 ∘C to 5 ∘C.
q3 = m3 * C3 * ΔT3
Now, let's put this all together:
q1 + q2 + q3 = 0
Since the heat transferred from the water and the heat transferred to the ice both contribute to lowering the temperature, the sum of these quantities should be zero.
From the given information, we know:
C1 = 4.18 J/g⋅∘C (specific heat capacity of water)
C3 = 2.09 J/g⋅∘C (specific heat capacity of ice)
Hf = 6.02 kJ/mol = 6.02 J/g (heat of fusion of ice)
Now, let's plug in the numbers and solve for the mass of the ice (m2):
m1 * C1 * ΔT1 + m2 * Hf + m3 * C3 * ΔT3 = 0
Since we want to find the mass of ice in grams, we need to convert 355 mL of water to grams:
m1 = 355 g (density of water is 1.0 g/mL)
We can also calculate the change in temperature (ΔT) for each step:
ΔT1 = 0 ∘C - 25 ∘C = -25 ∘C
ΔT3 = 5 ∘C - 0 ∘C = 5 ∘C
Now let's substitute the values into the equation:
355 g * 4.18 J/g⋅∘C * -25 ∘C + m2 * 6.02 J/g + m3 * 2.09 J/g⋅∘C * 5 ∘C = 0
Simplifying the equation:
-4692.5 + 6.02m2 + 10.45m3 = 0
Rearranging the equation:
6.02m2 + 10.45m3 = 4692.5
Since the volume of water produced by the melting ice is considered negligible, we can assume the final mass of the system (water + ice) remains constant. Therefore:
m1 + m2 + m3 = constant
355 g + m2 + 0 = constant
Therefore, we can rewrite the equation as:
6.02m2 + 10.45(355 - m2) = 4692.5
Now, solve the equation to find the mass of ice (m2).