Problem 107 - University Physics with Modern Physics

Question

A worker pours 1.250 kg of molten lead at a temperature of \(365.0^{\circ} \mathrm{C}\) into 0.5000 \(\mathrm{kg}\) of water at a temperature of \(75.00^{\circ} \mathrm{C}\) in an insulated bucket of negligible mass. Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium.

Step-by-Step Solution

Answer

The mass remaining is the lead left after solidification and water left after vaporization.

Step 1: Understand the Problem

We have a system with molten lead and water reaching thermal equilibrium inside an insulated bucket. We need to calculate how much lead will solidify and how much water will vaporize without heat loss to the surroundings.

Step 2: Determine the Heat Exchange

We'll use the principle of conservation of energy, where the heat lost by the lead will be equal to the heat gained by the water. The specific heat capacity of lead is 0.128 J/g°C, and the latent heat of fusion is 24.5 J/g. The specific heat capacity of water is 4.186 J/g°C, and the latent heat of vaporization is 2260 J/g.

Step 3: Calculate Heat Lost by Lead Cooling to 327°C

Lead changes from molten at 365°C to solid at its melting point 327°C. First, calculate the heat loss by cooling: \[ Q_{cooling} = m_{lead} \times c_{lead} \times (T_{initial} - T_{final}) = 1250 \times 0.128 \times (365 - 327) \]

Step 4: Calculate Heat Lost by Lead Solidifying at 327°C

Next, calculate the heat loss due to phase change (solidification): \[ Q_{solidfy} = m_{lead} \times L_{fusion} = 1250 \times 24.5 \]

Step 5: Calculate Heat Lost by Solid Lead Cooling to 100°C

Calculate the heat loss if the lead cools further from 327°C to 100°C: \[ Q_{solid\_cooling} = m_{lead} \times c_{lead} \times (327 - 100) \]

Step 6: Calculate Heat Gained by Water Heating to 100°C

Next, calculate the heat required to raise the temperature of the entire water mass from 75°C to 100°C:\[ Q_{heat\_water} = m_{water} \times c_{water} \times (T_{final} - T_{initial}) = 500 \times 4.186 \times (100 - 75) \]

Step 7: Calculate Heat Required for Water to Vaporize

Finally, determine the heat necessary to vaporize water:\[ Q_{vaporize} = m_{vaporized} \times L_{vaporization} \]

Key Concepts

Heat Transfer

Heat transfer is a fundamental process that involves the movement of thermal energy from one substance to another. In our exercise, heat transfer happens between molten lead and water, striving to reach a balanced energy state, known as thermal equilibrium. The direction of heat flow is always from the hotter object—in this case, molten lead—toward the cooler object, which is water. This flow continues until both substances have the same temperature.

There are three primary modes through which heat can be transferred: conduction, convection, and radiation. In our scenario, conduction is the primary form of heat transfer. Conduction occurs when heat moves through direct contact between the moleculer structure of the lead and the water. This process is beautifully governed by the principle that heat lost by the hotter body (lead) is exactly equal to the heat gained by the colder body (water).

Understanding heat transfer is crucial in calculations involving cooling and phase transitions, which are essential components of energy transfers like the solidifying of lead and the vaporizing of water. This methodical exchange of energy results in the cooling of lead and possible heating of water to its boiling point.

Phase Change

Phase change refers to the transition of a substance from one state of matter to another, such as from solid to liquid or liquid to gas. In the given exercise, we observe two main phase changes: solidification of lead and the vaporization of water.

When lead cools down from its molten state, it solidifies at its melting point, which is around 327°C. This phase transition from liquid to solid releases latent heat, known as the heat of fusion. For lead, this requires 24.5 J/g. Consequently, as the lead releases this energy, it aids in warming the water and may facilitate its transition into vapor.

If the water is heated up to its boiling point, it might vaporize, undergoing a phase change from liquid to gas. This process requires a substantial amount of energy, termed as the latent heat of vaporization, which is 2260 J/g for water. Whether or not the water vaporizes depends on how much total energy is transferred from the lead to the water.

Specific Heat Capacity

Specific heat capacity is an essential concept that defines how much energy is required to raise the temperature of a substance by one degree Celsius. It is a measure of how resistant a material is to changing its temperature. Both lead and water have specific heat capacities that dictate how they respond to heat transfer.

For lead, the specific heat capacity is 0.128 J/g°C. This relatively low value means lead heats up or cools down quickly with small amounts of added or removed heat. Conversely, water has a high specific heat capacity of 4.186 J/g°C, making it very effective at absorbing heat with only slight changes in temperature.

In our exercise, these properties play a critical role in determining how much each substance's temperature changes during heat exchange, which ultimately affects whether the lead solidifies and/or the water vaporizes.

• Lead quickly loses heat compared to water due to its lower specific heat capacity.
• Water requires more energy to increase its temperature than lead does for the same mass.

Conservation of Energy

The law of conservation of energy is a fundamental principle stating that energy in a closed system is constant unless acted on by external forces. This means that energy cannot be created or destroyed, only transformed from one form to another.

In our insulated system containing molten lead and water, the conservation of energy underlines the behavior of these materials as they reach thermal equilibrium. Since the bucket is insulated, no heat escapes to the surroundings. As a result, the energy lost by the lead as it cools must equal the energy gained by the water.

This energy balance allows us to calculate the extent of phase changes and temperature shifts in the given exercise. We apply this principle by equating the heat lost by the lead (both by cooling and solidifying) to the heat gained by the water (by heating and potentially vaporizing). It ensures that every joule lost by lead is accounted for by an equivalent energy gain in the water, maintaining energy conservation throughout the process.

• Energy lost by lead = Energy gained by water.
• No external energy influences the system, emphasizing internal energy exchanges.