Question

Suppose 100 . g of ice at \(0^{\circ} \mathrm{C}\) are added to 300 . \(\mathrm{g}\) of water at \(25^{\circ} \mathrm{C}\). Is this sufficient ice to lower the temperature of the system to \(0^{\circ} \mathrm{C}\) and still have ice remaining? Show evidence for your answer.

Short answer

Not all ice melts; some ice will remain.

Step-by-step solution

- Determine Heat Required to Melt Ice

Calculate the heat required to melt 100 g of ice using the formula: \[ Q = m \times L_f \]where \( Q \) is the heat, \( m \) is the mass of the ice, and \( L_f \) is the latent heat of fusion. For ice, \( L_f = 334\ \text{J/g} \). Thus: \[ Q = 100\text{ g} \times 334\text{ J/g} = 33400\text{ J} \]

- Calculate Heat Available from Water

Calculate the heat that 300 g of water at \(25^\text{C}\) can release to reach \(0^\text{C}\) using the formula: \[ Q = m \times c \times \triangle T \]where \( Q \) is the heat, \( m \) is the mass, \( c \) is the specific heat capacity of water (4.18 J/g°C), and \( \triangle T \) is the change in temperature. \[ Q = 300\text{ g} \times 4.18 \text{ J/g°C} \times 25\text{°C} = 31350\text{ J} \]

- Compare Heat Required to Heat Available

The heat required to melt the ice (33400 J) is more than the heat that can be released by the water (31350 J). Thus, melting all the ice requires more energy than the water can provide. Since not all the ice will melt, there will be ice remaining.

Key concepts

latent heat of fusion

Suppose you want to understand why ice behaves the way it does when mixed with water. A crucial concept here is the 'latent heat of fusion'. This is the amount of heat energy required to change a substance (in this case, ice) from a solid to a liquid at its melting point, without changing its temperature.

For ice, the latent heat of fusion is 334 joules per gram (J/g). This means that each gram of ice needs 334 joules of energy to melt. When you add 100 g of ice to water, you will need:

• Calculation: Use the formula: \( Q = m \times L_f \), where \( Q \) is the heat, \( m \) is the mass of the ice, and \( L_f \) is the latent heat of fusion.
  
• Example: For 100 g of ice: \[ Q = 100 \text{ g} \times 334 \text{ J/g} = 33400 \text{ J} \]

This means you need 33400 joules to completely melt the ice.

Understanding the latent heat of fusion helps explain why ice can cool water effectively. The process absorbs a lot of heat energy before the ice even starts raising the water temperature.

specific heat capacity

Next, let's dive into 'specific heat capacity'. This is the amount of heat required to change the temperature of one gram of a substance by 1°C (1 degree Celsius).

For water, the specific heat capacity is 4.18 J/g°C. Specific heat capacity helps us determine how much heat the water can absorb or release during a temperature change.

To find out how much heat 300 g of water can release as it cools from 25°C to 0°C, use the formula:
\( Q = m \times c \times \triangle T \)
where \( Q \) is the heat, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \triangle T \) is the change in temperature.

• Example Calculation: For 300 g of water cooling from 25°C to 0°C: \[ Q = 300 \text{ g} \times 4.18 \text{ J/g°C} \times 25 \text{°C} = 31350 \text{ J} \]

This means that 300 g of water can release 31350 joules as it cools to 0°C.

Knowing how specific heat capacity works is essential for various applications. It tells us how thermal energy flows and how substances respond to heating or cooling.

phase change

Let's talk about a 'phase change'. A phase change occurs when a substance transitions from one state of matter to another. Common phase changes include melting, freezing, boiling, and condensation.

In our example, we focus on melting, where ice (solid) turns into water (liquid). This happens at 0°C for ice. The critical factor during a phase change is that the substance absorbs or releases heat without a change in temperature.

For example:

• Melting ice: When you add heat to ice at 0°C, it doesn't immediately raise the temperature. Instead, the energy goes into breaking the bonds between the ice molecules, turning it into water.
• Freezing water: Conversely, if you remove heat from water at 0°C, it will turn into ice without a temperature drop until the phase change is complete.

This concept helps explain why melting ice in our water scenario remains at 0°C until all 100 g of ice receives enough energy to complete the phase change.

In our exercise, since the water only provided 31350 joules but 33400 joules were required, not all the ice melted. Thus, the temperature dropped to 0°C, and some ice remained.

Understanding phase changes reveals a lot about how energy transfers work in practical situations, such as cooking, refrigeration, and even weather phenomena!