Calculate the enthalpy change when 45 g of ice at -40.0 oC is heated to completely melt to liquid water at 0.0oC. (The heat of fusion for water is 333 J/g.)
To calculate the enthalpy change, we need to consider two steps:
Step 1: Heating the ice from -40.0 oC to 0.0 oC
Step 2: Melting the ice at 0.0 oC to liquid water at 0.0 oC
Let's start with Step 1:
1. First, we need to calculate the heat required to raise the temperature of the ice from -40.0 oC to 0.0 oC.
The specific heat capacity of ice is 2.09 J/g oC.
To calculate the heat energy (q) required for this step, we use the formula:
q = mass × specific heat capacity × temperature change
Given:
mass = 45 g
specific heat capacity = 2.09 J/g oC
temperature change = 0.0 oC - (-40.0 oC) = 40.0 oC
Plugging in the values:
q = 45 g × 2.09 J/g oC × 40.0 oC
= 3762 J
Therefore, it requires 3762 J of heat energy to raise the ice's temperature from -40.0 oC to 0.0 oC.
Moving on to Step 2:
2. We need to calculate the heat required to melt the ice at 0.0 oC into liquid water at 0.0 oC.
The heat of fusion for water is given as 333 J/g.
To calculate the heat energy (q) required for this step, we use the formula:
q = mass × heat of fusion
Given:
mass = 45 g
heat of fusion = 333 J/g
Plugging in the values:
q = 45 g × 333 J/g
= 14,985 J
Therefore, it requires 14,985 J of heat energy to melt 45 g of ice to liquid water at 0.0 oC.
To find the total enthalpy change, we add the heat energy for both steps:
Total enthalpy change = Heat energy required in Step 1 + Heat energy required in Step 2
= 3762 J + 14,985 J
= 18,747 J
So, the enthalpy change when 45 g of ice at -40.0 oC is heated to completely melt to liquid water at 0.0 oC is 18,747 J.