Problem 76 - College Physics

Question

A worker pours \(1.250 \mathrm{~kg}\) of molten lead at a temperature of \(327.3^{\circ} \mathrm{C}\) 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 of lead and water remaining is 1.250 kg of solid lead and 0.500 kg of water.

Step 1: Identify the heat transfer process

The problem involves a heat transfer between molten lead and water until they reach thermal equilibrium. The heat lost by the lead will be gained by the water since the system is insulated and has negligible mass.

Step 2: List the known values and properties

Lead initial temperature: \(327.3^{\circ} \mathrm{C}\), water initial temperature: \(75.00^{\circ} \mathrm{C}\), mass of lead: \(1.250 \mathrm{kg}\), mass of water: \(0.5000 \mathrm{kg}\). Specific heat capacity of lead \(c_{lead} = 128 \mathrm{~J/kg \cdot °C}\). Specific heat capacity of water \(c_{water} = 4186 \mathrm{~J/kg \cdot °C}\). Melting point of lead: \(327.5^{\circ} \mathrm{C}\). Latent heat of fusion of lead: \(24.7 \mathrm{~kJ/kg}\).

Step 3: Determine the thermal equilibrium temperature

Set up the energy balance equation: the heat lost by lead equals the heat gained by water. Assuming the final temperature \(T_f\) is below lead's melting point (since the capacity of water is higher and water is at lower temperature), use: \(m_{lead} \cdot c_{lead} \cdot (T_{init, lead} - T_f) = m_{water} \cdot c_{water} \cdot (T_f - T_{init, water})\).

Step 4: Simplify the energy equation and calculate the final temperature \(T_f\)

Solve for \(T_f\):\[1.25 \times 128(327.3 - T_f) = 0.5 \times 4186(T_f - 75)\]This equation simplifies down to:\[160T_f = 33400 + 58840\]Solving for \(T_f\), substitute and calculate.

Step 5: Evaluate energy equation conditions

Given the high heat capacity of water, solve the equation numerically or graphically if needed to find whether the lead cools below its melting point, assuming no significant phase change occurs at equilibrium:

Step 6: Calculate the effect

If \(T_f < 327.5^{\circ} \mathrm{C}\), then since no phase change happened \[ T_f = 81.55^{\circ} \mathrm{C}\] . This means all lead returns to solid state and water remains liquid.

Step 7: Final outcome

Add the masses of both, as there are no losses: The masses of both lead and water remain as initially. 1.250 kg of lead remains solidifying nonetheless.

Key Concepts

Latent Heat

Latent heat is the amount of energy required to change the phase of a substance without changing its temperature. This is important in our problem because it dictates how much energy is absorbed or released during the phase change.
Think of it like a hidden energy transfer. When a substance melts or solidifies, it either absorbs or releases latent heat. Lead, in this example, has a latent heat of fusion of 24.7 kJ/kg.
This means, for each kilogram of molten lead that solidifies, it releases 24.7 kJ of energy without a temperature change.

• The latent heat of fusion is crucial when considering if the lead will solidify as it cools.
• The system's thermal energy does not increase; instead, it is absorbed into another process (e.g., phase change).

In this problem, since the final temperature is 81.55°C, which is below the melting point of lead (327.5°C), latent heat is not a major player here as the lead doesn’t remain molten. Instead, it solidifies completely.

Specific Heat Capacity

Specific heat capacity is the amount of energy required to raise the temperature of a unit mass of a substance by one degree Celsius. In simpler terms, it measures how much a substance resists temperature change.
Consider the specifics for our materials:

• Lead has a specific heat capacity of 128 J/kg°C, which is relatively low. It heats up and cools down quickly.
• Water, with a specific heat of 4186 J/kg°C, requires a lot more energy to change temperature. It is much more resistant to thermal changes.

In the context of our experiment, due to these differences, the thermal response of lead compared to water is vastly different.
The water's higher specific heat capacity contributes to its significant ability to absorb energy transferred from lead, slowing its temperature rise and preventing any significant temperature spike.

Thermal Equilibrium

Thermal equilibrium is achieved when two substances reach the same temperature and no more heat flows between them. This concept is at the heart of the problem since we're trying to find the temperature both the lead and water end up at.
An important thought to keep in mind is:

• When thermal equilibrium is reached, the energy gained by one substance must equal the energy lost by the other, providing the system is closed and insulated.
• The equation derived in this problem ensures the systems reach equilibrium without being impacted by external factors.

In our scenario, once the thermal equilibrium is achieved at approximately 81.55°C, there is no further heat transfer, and both the lead and the water temperatures stabilize.

Energy Balance

Energy balance in heat transfer refers to the idea that the total energy within an insulated system must remain constant.
For our exercise, it holds a vital role, especially since we are calculating temperature equilibrium.

• The energy lost by the lead as it cools equals the energy gained by the water as it heats up.
• This principle relies heavily on accurate calculations of specific heat capacities and initial temperatures, ensuring energy is neither lost nor gained from the outside.

For accurately determining the final temperature, the initial conditions laid out – such as masses and temperatures – help define the energy balance equation used.
In this case, the energy exchanges are perfectly balanced because the system is isolated (no heat loss to the surroundings), highlighting a critical aspect of energy conservation.

Phase Change

A phase change occurs when a substance transitions from one state of matter to another, such as solid to liquid or liquid to gas.
In the scenario with lead and water, the lead transitions from molten (liquid) back to solid as it cools to a final temperature of 81.55°C.

• The molten lead starts above its melting point and solidifies as the system cools and stabilizes.
• No phase change occurs in water, as its temperature never exceeds its boiling point nor drops to freezing.

This lack of significant phase change in water simplifies the calculations, focusing the phase change considerations mainly on the lead.
Understanding phase changes and their associated latent heats ensures accurate predictions of system behavior during thermal processes.