To find the energy needed to convert ice to water, we need to consider two steps:
Step 1: Heating the ice from -14.6°C to 0°C
Step 2: Melting the ice at 0°C to water at 25.8°C
Step 1: Heating the ice from -14.6°C to 0°C:
We will use the formula:
q1 = m * c * ΔT
Where:
q1 is the energy required,
m is the mass of the ice (104g),
c is the specific heat of ice at -14.6°C (2.01 J/g°C),
ΔT is the change in temperature (0°C - (-14.6°C)).
Substituting the values, we get:
q1 = 104g * 2.01 J/g°C * (0°C - (-14.6°C))
q1 = 104g * 2.01 J/g°C * 14.6°C
q1 ≈ 3027 J
Step 2: Melting the ice at 0°C to water at 25.8°C:
We need to account for the energy required to melt the ice at 0°C. The heat of fusion for ice is 334 J/g. We can calculate the energy needed using the formula:
q2 = m * ΔHf
Where:
q2 is the energy required,
m is the mass of the ice (104g),
ΔHf is the heat of fusion (334 J/g).
Substituting the values, we get:
q2 = 104g * 334 J/g
q2 ≈ 34,736 J
The total energy needed is the sum of q1 and q2:
Total energy = q1 + q2
Total energy ≈ 3027 J + 34,736 J
Total energy ≈ 37,763 J
Therefore, approximately 37,763 joules (or 37.8 kilojoules) of energy are needed to convert 104 g of ice at -14.6°C to water at 25.8°C.