Calculate ∆H for the constant pressure process of turning 1 mole
of ice at 0°C to liquid water at 15°C. CP of water is 4.18 J/g∙K,
Heat of fusion of ice at 0°C is 80.0 cal/g
To calculate the ∆H for the constant pressure process of turning 1 mole of ice at 0°C to liquid water at 15°C, we need to consider two steps: the heat required to raise the temperature of the ice from 0°C to 0°C (melting) and the heat required to raise the temperature of the resulting liquid water from 0°C to 15°C.
1. Heat required for melting ice at 0°C: This can be calculated using the heat of fusion. Given that the heat of fusion of ice at 0°C is 80.0 cal/g, we need to convert this to Joules and multiply it by the number of grams of ice.
1 mole of ice has a molar mass of approximately 18.015 g/mol, so 1 mole of ice is equal to 18.015 g of ice.
The heat required to melt 1 mole of ice is:
Q1 = (80.0 cal/g) x (1 cal/4.184 J) x (18.015 g) = 3053.52 J
2. Heat required to raise the temperature of liquid water from 0°C to 15°C: We need to know the heat capacity of water to calculate this. The specific heat capacity (CP) of water is given as 4.18 J/g∙K.
To heat the water from 0°C to 15°C, we need to multiply the CP by the mass of the water. The molar mass of water is approximately 18.015 g/mol, so we have 18.015 g of water.
The heat required to raise the temperature of 18.015 g of water by 15°C is:
Q2 = (4.18 J/g∙K) x (18.015 g) x (15 K) = 1126.79 J
To calculate the total heat change (∆H), we add the heat required for the two steps:
∆H = Q1 + Q2 = 3053.52 J + 1126.79 J = 4180.31 J
Therefore, the ∆H for the constant pressure process of turning 1 mole of ice at 0°C to liquid water at 15°C is approximately 4180.31 J.