Calculate the enthalpy change when 50. g of ice at -30.0 °C is heated to completely melt to liquid water at 0.0°C. (The heat of fusion for water is 333 J/g.)
q1 = heat required to raise T of ice from -30 to O C.
q1 = mass of ice x specific heat ice x (Tf - Ti) where Tf is final T (O) and Ti is initial T (-30).
q2 = heat required to melt the ice at zero; i.e., change from solid phase to liquid phase.
q2 = mass of ice x heat fusion.
total q = q1 + q2.
To calculate the enthalpy change when ice is heated and completely melted to liquid water, we need to consider two steps:
Step 1: Heating the ice to its melting point.
Step 2: Melting the ice into liquid water.
Let's calculate each step:
Step 1: Heating the ice to 0.0°C
To calculate the heat absorbed in this step, we can use the specific heat capacity equation:
q1 = m * C * ΔT
Where:
- q1 is the heat absorbed
- m is the mass of the ice (50.0 g)
- C is the specific heat capacity of ice (2.09 J/g°C)
- ΔT is the change in temperature (final temperature - initial temperature)
ΔT = 0.0°C - (-30.0°C) = 30.0°C
Substituting these values into the equation:
q1 = 50.0 g * 2.09 J/g°C * 30.0°C
q1 = 3135 J
Therefore, the heat absorbed in Step 1 is 3135 J.
Step 2: Melting the ice
The enthalpy change during melting is given by the heat of fusion equation:
q2 = m * ΔHfus
Where:
- q2 is the enthalpy change (heat of fusion)
- m is the mass of the ice (50.0 g)
- ΔHfus is the heat of fusion for water (333 J/g)
Substituting these values into the equation:
q2 = 50.0 g * 333 J/g
q2 = 16,650 J
Therefore, the enthalpy change during melting (Step 2) is 16,650 J.
The total enthalpy change is the sum of the heat absorbed in Step 1 and the enthalpy change in Step 2:
Total enthalpy change = q1 + q2
Total enthalpy change = 3135 J + 16,650 J
Total enthalpy change = 19,785 J
Therefore, the enthalpy change when 50.0 g of ice at -30.0°C is heated to completely melt into liquid water at 0.0°C is 19,785 J.
To calculate the enthalpy change when ice at -30.0 °C is heated to completely melt to liquid water at 0.0 °C, we need to consider two steps:
1. Heating the ice from -30.0 °C to 0.0 °C.
2. Melting the ice at 0.0 °C to liquid water at 0.0 °C.
Step 1: Heating the ice from -30.0 °C to 0.0 °C
To raise the temperature of the ice from -30.0 °C to 0.0 °C, we need to use the specific heat capacity of ice, which is 2.09 J/g °C.
The formula to calculate the heat energy is Q = m * c * ΔT, where:
Q = heat energy (enthalpy change)
m = mass of the substance (in grams)
c = specific heat capacity of the substance
ΔT = change in temperature
Using the given values:
m = 50.0 g
c = 2.09 J/g °C
ΔT = (0.0 °C) - (-30.0 °C) = 30.0 °C
Q = (50.0 g) * (2.09 J/g °C) * (30.0 °C)
Q = 3135 J
So, the heat energy required to heat the ice from -30.0 °C to 0.0 °C is 3135 J.
Step 2: Melting the ice at 0.0 °C to liquid water at 0.0 °C
The heat of fusion for water is given as 333 J/g. This tells us that 333 J of heat energy is required to melt 1 gram of ice.
Since we have 50.0 g of ice, we can calculate the heat energy required using the formula:
Q = m * ΔH_fusion
Q = (50.0 g) * (333 J/g)
Q = 16650 J
So, the heat energy required to melt the ice at 0.0 °C to liquid water at 0.0 °C is 16650 J.
Therefore, the total enthalpy change is the sum of the heat energy required in step 1 and step 2:
Total enthalpy change = 3135 J + 16650 J
Total enthalpy change = 19785 J
Hence, the enthalpy change when 50.0 g of ice at -30.0 °C is heated to completely melt to liquid water at 0.0 °C is 19785 J.