Problem 57 - Physics

Question

A woman finds the front windshield of her car covered with ice at \(-12.0^{\circ} \mathrm{C}\). The ice has a thickness of \(4.50 \times 10^{-4} \mathrm{~m},\) and the windshield has an area of \(1.25 \mathrm{~m}^{2}\). The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\). How much heat is required to melt the ice?

Step-by-Step Solution

Answer

The heat required to melt the ice is 172,010 J.

Step 1: Calculate the Mass of the Ice

First, calculate the mass of the ice using the formula \( \text{mass} = \text{density} \times \text{volume} \). The volume of the ice can be found using \( \text{volume} = \text{area} \times \text{thickness} \). So, \( \text{volume} = 1.25 \, \mathrm{m}^2 \times 4.50 \times 10^{-4} \, \mathrm{m} = 5.625 \times 10^{-4} \, \mathrm{m}^3 \). The mass is \( 917 \, \mathrm{kg/m}^3 \times 5.625 \times 10^{-4} \, \mathrm{m}^3 = 0.515 \mathrm{kg} \).

Step 2: Determine the Heat Required to Melt the Ice

The heat required to melt the ice can be calculated using the formula \( Q = m \times L_f \), where \( m \) is the mass of the ice and \( L_f \) is the latent heat of fusion for ice (\( 334,000 \, \mathrm{J/kg} \)). Substitute the values into the equation to obtain \( Q = 0.515 \, \mathrm{kg} \times 334,000 \, \mathrm{J/kg} = 172,010 \, \mathrm{J} \).

Key Concepts

Heat Transfer

Heat transfer is the process of energy moving from one place to another due to temperature differences. In our exercise, heat is needed to melt ice on a windshield, demonstrating a practical application of heat transfer. When you add heat to a cold object like ice, it absorbs energy, causing it to change state from solid to liquid. This is an essential concept in thermodynamics, particularly when considering how energy moves through different states of matter.

To understand this concept better, consider these points:

• Heat naturally flows from warmer areas to cooler areas.
• When a substance absorbs heat, its temperature may rise or it may change state.
• In the exercise, the heat is changing the state of the ice without raising its temperature until it becomes liquid.

Latent Heat

Latent heat refers to the energy required to change a substance's state. For example, melting ice into water requires latent heat of fusion. This is essential to know when melting substances, as it calculates how much energy is needed to change their state.

The latent heat of fusion for ice is a constant value:

• For ice, it is about 334,000 J/kg. This is the amount of energy needed to melt 1 kg of ice.

In our problem, we use this value to find out how much energy is necessary to melt the ice covering a windshield. The formula used is:

• \( Q = m \times L_f \) where \( Q \) is heat, \( m \) is mass, and \( L_f \) is the latent heat of fusion.

This energy does not cause a temperature change but instead powers the transformation of ice into liquid.

Mass Calculation

Mass calculation involves using the relationship between density, volume, and mass. In the exercise, we're trying to find out how much ice is on the windshield.

Here's how we determine the mass:

• First, calculate the volume using the ice's thickness and the windshield’s area: \( \text{volume} = \text{area} \times \text{thickness} \).
• Substitute the values to get \( 1.25 \, \text{m}^2 \times 4.50 \times 10^{-4} \, \text{m} = 5.625 \times 10^{-4} \, \text{m}^3 \).
• Next, use the density of ice \( (917 \, \text{kg/m}^3) \) to find the mass: \( \text{mass} = \text{density} \times \text{volume} \).
• The mass comes out to be \( 0.515 \, \text{kg} \).

Density of Ice

The density of ice is a key factor in our calculations. It helps in determining how much a given volume of ice will weigh. Density is defined as mass per unit volume, and for ice, it is typically \( 917 \, \text{kg/m}^3 \).

Knowing the density allows us to:

• Convert volumes to masses, which is crucial when quantifying materials in thermodynamic problems.
• Apply this information in the formula \( \text{mass} = \text{density} \times \text{volume} \).

Understanding the density of ice can help you grasp why ice floats on water and how much material you're dealing with in both scientific and everyday scenarios.