The ice skate of a 210-pound hockey player is in contact with the ice over a 2.5mm x 15cm area. Determine the freezing point of the ice under the blade of the skate?
To determine the freezing point of the ice under the blade of the skate, we need to understand how pressure affects the freezing point of water.
Water typically freezes at 0 degrees Celsius (32 degrees Fahrenheit) under normal atmospheric pressure. However, when pressure is applied to the ice, such as in this case with the skate blade, the freezing point of water is lowered.
The relationship between pressure and the freezing point of water can be described by the concept of freezing point depression. According to this concept, the freezing point of a substance is lowered by a certain amount for every unit of pressure applied.
To calculate the freezing point of the ice under the blade of the skate, we need to use the formula for freezing point depression:
ΔT = K * m
Where:
ΔT is the change in freezing point
K is the cryoscopic constant (which is dependent on the solvent, in this case water)
m is the molality of the solute (which is the concentration of the solute in moles per kilogram of solvent)
In this case, the solute is the pressure applied by the hockey player's weight, and the solvent is water.
To calculate the molality (m), we need to convert the given weight of the player from pounds to kilograms:
Weight = 210 pounds
1 pound = 0.4536 kilograms
Weight = 210 * 0.4536 = 95.472 kilograms
Now, let's calculate the pressure applied by the player's weight:
Area = 2.5mm x 15cm = 0.0025m x 0.15m = 0.000375 square meters
Pressure = Force / Area
Force = Weight * Acceleration due to gravity
Acceleration due to gravity = 9.8 m/s² (approximate value on Earth)
Force = 95.472 kg * 9.8 m/s² = 936.8256 N
Pressure = 936.8256 N / 0.000375 square meters = 2495674.93 Pa
Note: Pa represents Pascal, which is the SI unit for pressure.
Now, we need to find the cryoscopic constant (K) for water. The cryoscopic constant for water is approximately 1.853 K * kg/mol.
Now we can substitute the values into the formula:
ΔT = K * m
ΔT = 1.853 K * kg/mol * (2495674.93 Pa / (18 g/mol) / 0.375 kg)
Since the molality (m) is the moles of solute per kilogram of solvent, we need to find out the number of moles in the given pressure.
The molar mass of water (H2O) is approximately 18 g/mol, so:
m = 2495674.93 Pa / (18 g/mol) / 0.375 kg
Calculating m:
m = 2495674.93 Pa / (18 g/mol) / 0.375 kg
m ≈ 72908.52 mol/kg
Finally, substitute the values:
ΔT = 1.853 K * kg/mol * 72908.52 mol/kg
ΔT ≈ 135253.97 K
Therefore, the freezing point of the ice under the blade of the skate is lowered by approximately 135,253.97 Kelvin. To determine the actual freezing point, subtract this value from the standard freezing point of water, which is 0 degrees Celsius (273.15 Kelvin).
Actual freezing point = 0°C - 135253.97 K = - 135253.97 K