How much heat energy, in kilojoules, is required to convert 63.0g of ice at −18.0 ∘C to water at 25.0 ∘C ?

To calculate the heat energy required to convert the ice to water, we need to consider two separate processes: the heat energy required to increase the temperature of the ice from -18.0 °C to 0 °C, and the heat energy required to convert the ice at 0 °C to water at 25.0 °C.

First, let's calculate the heat energy required to increase the temperature of the ice from -18.0 °C to 0 °C. The specific heat capacity of ice is 2.09 J/g°C.

The mass of the ice is 63.0 g. The temperature change is:
ΔT = 0 °C - (-18.0 °C) = 18.0 °C

The heat energy required for this process is:
Q1 = mass × specific heat capacity × ΔT
= 63.0 g × 2.09 J/g°C × 18.0 °C

Converting the result from joules to kilojoules:
Q1 = (63.0 g × 2.09 J/g°C × 18.0 °C) / 1000
= 2.19894 kJ

Now, let's calculate the heat energy required to convert the ice at 0 °C to water at 25.0 °C. The enthalpy of fusion for ice is 334 J/g.

The mass of the ice is still 63.0 g.

The heat energy required for this process is:
Q2 = enthalpy of fusion × mass

Converting the result from joules to kilojoules:
Q2 = (334 J/g × 63.0 g) / 1000
= 21.042 kJ

Finally, we can calculate the total heat energy required by summing up Q1 and Q2:
Total heat energy = Q1 + Q2
= 2.19894 kJ + 21.042 kJ
= 23.24094 kJ

Therefore, the total amount of heat energy required to convert 63.0g of ice at -18.0 °C to water at 25.0 °C is approximately 23.24 kJ.

To calculate the heat energy required to convert ice to water, you can use the following formula:

q = m * ΔHfus + q1 + q2

where:
q is the total heat energy,
m is the mass of the substance,
ΔHfus is the heat of fusion,
q1 is the heat energy required to raise the temperature of the ice to its melting point, and
q2 is the heat energy required to raise the temperature of the water from its melting point to the final temperature.

Let's break down the calculations step by step:

Step 1: Calculate q1 - the heat energy required to raise the temperature of the ice to its melting point (0°C):
q1 = m * c * ΔT1

In this case, the mass (m) is 63.0g, the specific heat capacity for ice (c) is 2.09 J/g°C, and the temperature change (ΔT1) is from -18.0°C to 0°C.

q1 = 63.0g * 2.09 J/g°C * (0°C - (-18.0°C))

Step 2: Calculate q2 - the heat energy required to raise the temperature of the water from its melting point to the final temperature (25°C):
q2 = m * c * ΔT2

In this case, the mass (m) is still 63.0g, the specific heat capacity for water (c) is 4.18 J/g°C, and the temperature change (ΔT2) is from 0°C to 25°C.

q2 = 63.0g * 4.18 J/g°C * (25.0°C - 0°C)

Step 3: Calculate ΔHfus - the heat of fusion for ice:
ΔHfus = m * Hfus

In this case, the mass (m) is still 63.0g, and the heat of fusion for water (Hfus) is 334 J/g.

ΔHfus = 63.0g * 334 J/g

Step 4: Sum up q1, q2, and ΔHfus to get the total heat energy:
q = q1 + q2 + ΔHfus

Now, let's calculate the actual values:

Step 1: q1 = 63.0g * 2.09 J/g°C * (0°C - (-18.0°C))
= 63.0g * 2.09 J/g°C * 18.0°C

Step 2: q2 = 63.0g * 4.18 J/g°C * (25.0°C - 0°C)
= 63.0g * 4.18 J/g°C * 25.0°C

Step 3: ΔHfus = 63.0g * 334 J/g

Step 4: q = q1 + q2 + ΔHfus

Finally, convert the total heat energy (q) from joules to kilojoules by dividing by 1000:

q = q / 1000

Now, you can perform these calculations to find the exact value of the heat energy required to convert 63.0g of ice at -18.0°C to water at 25.0°C.

You work this in stages.

Within a phase it is
q = mass x specific heat x (Tfinal-Tinitial). For example, for liquid water at zero C to boiling it will be
q = mass H2O x specific heat liquid H2O x (100-0).
Then at a phase change (from ice to liquid or from liquid to gas it is
q = mass x heat fusion (for solid) or heat vaporization (for liquid to gas).
Then add the parts to find the total.

234