Question

Calculate the quantity of heat required to convert \(60.1 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) at \(0.0^{\circ} \mathrm{C}\) to \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100.0^{\circ} \mathrm{C} .\) The heat of fusion of ice at \(0^{\circ} \mathrm{C}\) is \(333 \mathrm{J} / \mathrm{g}\); the heat of vaporization of liquid water at \(100^{\circ} \mathrm{C}\) is \(2260 \mathrm{J} / \mathrm{g}\).

Short answer

Total heat required is 180951.3 J.

Step-by-step solution

Calculate heat for melting ice

To calculate the heat required to melt 60.1 g of ice at 0°C into liquid water at 0°C, use the formula: \( q = m \cdot \Delta H_f \), where \( m \) is the mass and \( \Delta H_f \) is the heat of fusion. Here, \( m = 60.1 \text{ g} \) and \( \Delta H_f = 333 \text{ J/g} \). So, \( q = 60.1 \text{ g} \times 333 \text{ J/g} = 20019.3 \text{ J} \).

Calculate heat for heating water

Next, calculate the heat required to raise the temperature of the liquid water from 0°C to 100°C. Use the formula: \( q = m \cdot c \cdot \Delta T \), where \( c = 4.18 \text{ J/g°C} \) for water, \( m = 60.1 \text{ g} \), and \( \Delta T = 100°C \). Thus, \( q = 60.1 \text{ g} \times 4.18 \text{ J/g°C} \times 100°C = 25106 \text{ J} \).

Calculate heat for vaporization

Now, calculate the heat required to vaporize the liquid water at 100°C into vapor at 100°C. Use the formula: \( q = m \cdot \Delta H_v \), where \( \Delta H_v = 2260 \text{ J/g} \). Thus, \( q = 60.1 \text{ g} \times 2260 \text{ J/g} = 135826 \text{ J} \).

Sum all heat quantities

Add up all the calculated heat quantities to find the total heat required: \( q_{\text{total}} = 20019.3 \text{ J} + 25106 \text{ J} + 135826 \text{ J} = 180951.3 \text{ J} \).

Key concepts

Heat of Fusion

The heat of fusion represents the energy required for a solid to convert into a liquid without changing its temperature. Specifically for water, the heat of fusion is the energy needed to transform ice at 0°C into liquid water at the same temperature.

• For ice, the heat of fusion is about 333 J/g, meaning 333 joules of heat energy is required to melt 1 gram of ice.
• This process occurs without a temperature change, as all the energy goes into breaking the bonds holding the ice molecules in a solid structure.

In our problem, we're converting 60.1 grams of ice into water. Thus, using the formula \( q = m \cdot \Delta H_f \), we multiply the mass of the ice by the heat of fusion to find the total energy required for this phase change. Remember, the temperature remains steady at 0°C change during the whole process.

Heat of Vaporization

The heat of vaporization is the energy required to change a liquid into a gas without altering its temperature. For water, this is needed when boiling water at 100°C into steam.

• The heat of vaporization for water is approximately 2260 J/g. This is the energy needed to convert 1 gram of liquid water at 100°C to steam at 100°C.
• This phase transition requires more energy compared to melting due to the higher bond strength between liquid molecules compared to solid.

In our exercise, we vaporize 60.1 grams of water. By applying the formula \( q = m \cdot \Delta H_v \), the total energy necessary to achieve this vaporization is calculated by multiplying the water's mass with its heat of vaporization, with the temperature holding steady at 100°C.

Specific Heat Capacity

Specific heat capacity is a property that tells us how much heat energy is needed to change the temperature of a substance. For water, it is 4.18 J/g°C.

• This value indicates that 4.18 joules of heat are required to raise the temperature of 1 gram of water by 1°C.
• The specific heat capacity plays a key role in the heating step between two phase changes—like warming water from 0°C to 100°C.

In our scenario, the specific heat capacity of water helps us compute the energy needed to heat 60.1 grams of water from freezing to boiling. By using the formula \( q = m \cdot c \cdot \Delta T \), where \( \Delta T \) is the rise in temperature (100°C in this case), you determine the energy required for this leg of the journey. The result showcases the considerable amount of heat required before entering the vaporization phase.