Question

The molar heats of fusion and vaporization for water are 6.02 \(\mathrm{kJ} / \mathrm{mol}\) and \(40.6 \mathrm{~kJ} / \mathrm{mol},\) respectively, and the specific heat capacity of liquid water is \(4.18 \mathrm{~J} / \mathrm{g}\). \({ }^{\circ} \mathrm{C}\). What quantity of heat energy is required to melt \(25.0 { }^{\circ} \mathrm{C}\) of ice at 0 \({ }^{\circ} \mathrm{C}\) ? What quantity of heat is required to vaporize \(37.5 \mathrm{~g}\) of liquid water at \(100 \) \({ }^{\circ} \mathrm{C}\) \(?\) What quantity of heat is required to warm \(55.2 \mathrm{~g}\) of liquid water from 0 ' \(\mathrm{C}\) to \(100 .\) \({ }^{\circ} \mathrm{C}\) ?

Short answer

8.353 kJ of heat energy is required to melt 25°C of ice at 0°C, 84.487 kJ of heat energy is required to vaporize 37.5 g of liquid water at 100°C, and 23.075 kJ of heat energy is required to warm 55.2 g of liquid water from 0°C to 100°C.

Step-by-step solution

Scenario 1: Melting ice at 0°C

First, we need to find the number of moles (n) of ice: n = mass / molar_mass. The mass of ice is \(25 g\) and the molar mass of water = 18.015 g/mol. Therefore, n = 25 g / 18.015 g/mol ≈ 1.387 mol Now, to find the heat energy needed to melt the ice: q = n * ΔH_fusion = 1.387 mol * 6.02 kJ/mol = 8.353 kJ Thus, 8.353 kJ of heat energy is required to melt 25°C of ice at 0°C.

Scenario 2: Vaporizing liquid water at 100°C

Find the number of moles (n) of liquid water: n = mass / molar_mass. The mass of liquid water is \(37.5 g\) and the molar mass of water = 18.015 g/mol. Therefore, n = 37.5 g / 18.015 g/mol ≈ 2.081 mol Now, to find the heat energy needed to vaporize the water: q = n * ΔH_vaporization = 2.081 mol * 40.6 kJ/mol = 84.487 kJ Thus, 84.487 kJ of heat energy is required to vaporize 37.5 g of liquid water at 100°C.

Scenario 3: Warming liquid water from 0°C to 100°C

To find the heat energy needed to warm the liquid water, we need the mass, specific heat capacity, and the change in temperature (ΔT). The mass of liquid water is \(55.2 g\), specific heat capacity (c) is \(4.18 J/g°C\), and the change in temperature (ΔT) is (100 - 0) °C = 100°C. Now, using the formula q = m * c * ΔT q = 55.2 g * 4.18 J/g°C * 100°C = 23074.72 J We can convert this value to kJ: 23074.72 J * (1 kJ / 1000 J) = 23.075 kJ Thus, 23.075 kJ of heat energy is required to warm 55.2 g of liquid water from 0°C to 100°C.

Key concepts

Molar Heat of Fusion

When we talk about changing a substance from solid to liquid, like ice melting into water, we use a term called molar heat of fusion. This refers to the amount of heat energy required to melt one mole of a solid at its melting point without changing its temperature. For water, this energy is 6.02 \(\mathrm{kJ/mol}\).

Let's break that down further. If you have a given mass of ice, to find out how much heat is needed to melt it completely, you first need to calculate how many moles of ice you have. You can find this by dividing the mass of the ice by the molar mass of water (18.015 g/mol). Once you know the number of moles, you multiply that by the molar heat of fusion.

For example, for 25 g of ice, converting its mass to moles would give you about 1.387 moles of ice. Multiplying this by 6.02 \(\mathrm{kJ/mol}\) reveals that approximately 8.353 kJ of energy is needed to melt all the ice at its melting temperature of 0°C.

Molar Heat of Vaporization

The molar heat of vaporization is a concept used when a liquid, like water, turns into a gas, or vapor. It's the amount of heat needed to turn one mole of a liquid at its boiling point into a gas with no temperature change.

For water, this value stands at 40.6 \(\mathrm{kJ/mol}\). To find the energy required to vaporize any given mass of water, calculate the moles of water just like you did with ice. Divide the mass of your water sample by the molar mass of water. For instance, if you have 37.5 g of water, that equates to about 2.081 moles.

You then multiply the number of moles by 40.6 \(\mathrm{kJ/mol}\) which tells you the total energy needed. For 37.5 g of water, it would take approximately 84.487 kJ to completely turn it into vapor at 100°C.

Specific Heat Capacity

Specific heat capacity is like a measure of how much heat energy a certain amount of a substance can absorb before it raises its temperature by 1 degree Celsius. For water, it's 4.18 \(\mathrm{J/g°C}\). This means to increase the temperature of one gram of water by one degree Celsius, you need 4.18 Joules of energy.

Calculating the amount of heat necessary to change the temperature of a larger sample of water involves multiplying the mass of the water by its specific heat capacity and the temperature change (ΔT).

For instance, if you need to heat 55.2 g of water from 0°C to 100°C, you first determine the ΔT, which is 100°C. Using the formula \(q = m \times c \times ΔT\), you would calculate \(55.2 \times 4.18 \times 100\), resulting in 23,074.72 J. Converting that into kilojoules (since 1 kJ = 1000 J), you find it takes about 23.075 kJ to heat the water to 100°C.