The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol • K. The conversion of 50.0 g of ice at 0.00°C to liquid water at 22.0°C requires how much kJ of heat?
Can someone explain how to do this, I tried but keep getting the wrong answer.
21.3 kJ of heat.
no explanation sorry im tkaing a test and i just wanted ot help i found out this was the anwser lmao
To calculate the amount of heat required to convert ice at 0.00°C to liquid water at 22.0°C, you need to consider two steps:
Step 1: Calculate the heat required to raise the temperature of ice from 0.00°C to 0.00°C (melting point).
Step 2: Calculate the heat required to raise the temperature of liquid water from 0.00°C to 22.0°C.
Step 1: Calculate the heat required to melt the ice at 0.00°C:
To raise the temperature of ice from 0.00°C to its melting point (0.00°C), no phase change occurs, so we only need to consider the heat capacity of ice.
The enthalpy change (∆H) can be calculated using the formula:
∆H = mass × heat capacity × ∆T
Given:
Mass of ice (m) = 50.0 g
Heat capacity of ice (C) = 75.3 J/mol • K
Temperature change (∆T) = (0 - 0)°C = 0 K
Since the heat capacity is given in J/mol • K, we need to convert the mass of ice (g) to moles by dividing by the molar mass of water:
Molar mass of water (H₂O) = 18.015 g/mol
Number of moles of water (n) = mass (m) / molar mass (M)
n = 50.0 g / 18.015 g/mol ≈ 2.7744 mol
∆H1 = n × C × ∆T
∆H1 = 2.7744 mol × 75.3 J/mol • K × 0 K = 0 J
Step 2: Calculate the heat required to raise the temperature of liquid water from 0.00°C to 22.0°C:
To raise the temperature of liquid water from 0.00°C to 22.0°C, we need to consider the specific heat capacity of liquid water.
The enthalpy change (∆H) can be calculated using the formula:
∆H = mass × specific heat capacity × ∆T
Given:
Mass of water (m) = 50.0 g
Specific heat capacity of liquid water (c) = 4.18 J/g • K
Temperature change (∆T) = (22.0 - 0)°C = 22.0 K
∆H2 = m × c × ∆T
∆H2 = 50.0 g × 4.18 J/g • K × 22.0 K
Finally, to calculate the total heat required (∆Htotal), add the enthalpy changes (∆H1 and ∆H2) together:
∆Htotal = ∆H1 + ∆H2
Remember that 1 kJ = 1000 J, so divide the final answer by 1000 to convert it to kilojoules.
I hope this explanation helps! Let me know if you have any further questions.
50/18.015 = moles in 50 g H2O.
q1 = heat to melt ice.
q1 = moles H2O x 6.01 kJ/mol
q2 = heat to move ice water from zero C to 22 C.
q2 = moles water x 75.3 J/mole x (22-0).
Change q1 to J and add to q2.