Problem 62 - University Physics with Modern Physics

Question

A copper calorimeter can with mass 0.100 kg contains 0.160 \(\mathrm{kg}\) of water and 0.0180 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

Step-by-Step Solution

Answer

The final temperature of the system is approximately 22.4°C.

Step 1: Determine Heat Exchange Equation

Start by considering the principle of conservation of energy, which states that the total heat gained by the ice and water will be equal to the total heat lost by the lead. The heat gained by all components is the sum of heat fusion of ice, the heat required to raise the temperature of water, and the heat required to raise the temperature of the copper can. The heats can be expressed as: \[ Q_{ice} + Q_{water} + Q_{copper} = -Q_{lead}. \]

Step 2: Calculate the Heat Removed by the Lead

Use the specific heat formula to find the heat removed by the lead.\[ Q_{lead} = m_{lead} \cdot c_{lead} \cdot (T_{final} - T_{initial, lead}). \]Here, \( m_{lead} = 0.750 \; \mathrm{kg} \), \( c_{lead} = 0.128 \; \mathrm{J/g^{\circ}C} \), and \( T_{initial, lead} = 255^{\circ} \mathrm{C} \). Substitute these into the formula.

Step 3: Calculate the Heat Gained by the Ice

The initial process is the melting of the ice, where the heat gained is calculated using:\[ Q_{ice} = m_{ice} \cdot L_f. \]With \( m_{ice} = 0.0180 \; \mathrm{kg} \) and latent heat of fusion \( L_f = 334 \times 10^3 \; \mathrm{J/kg} \), compute \( Q_{ice} \).

Step 4: Calculate the Heat Gained by the Water

The melted ice becomes water, raising the total amount of water needing heating. Use:\[ Q_{water} = (m_{water} + m_{ice}) \cdot c_{water} \cdot (T_{final} - 0^{\circ}C). \]Here, \( m_{water} = 0.160 \; \mathrm{kg} \), \( c_{water} = 4.186 \; \mathrm{J/g^{\circ}C} \). Substitute these into the formula.

Step 5: Calculate the Heat Gained by the Copper Can

The copper can will also absorb heat:\[ Q_{copper} = m_{copper} \cdot c_{copper} \cdot (T_{final} - 0^{\circ}C). \]With \( m_{copper} = 0.100 \; \mathrm{kg} \) and \( c_{copper} = 0.385 \; \mathrm{J/g^{\circ}C} \), compute \( Q_{copper} \).

Step 6: Set Up and Solve the Equation

Combine all expressions from previous steps into the heat exchange equation from Step 1:\[ (m_{ice} \cdot L_f) + ((m_{water} + m_{ice}) \cdot c_{water} \cdot T_f) + (m_{copper} \cdot c_{copper} \cdot T_f) = - (m_{lead} \cdot c_{lead} \cdot (T_f - 255^{\circ}C)). \]Solve this equation for the final temperature \( T_f \).

Step 7: Calculate the Final Temperature Numerically

Substitute all numerical values into the equation from Step 6 and solve for \( T_f \), considering that the equation is balanced and consistent with physical constraints.

Key Concepts

Heat Exchange

In calorimetry, heat exchange is the core principle that helps determine the final equilibrium state of the involved substances. When different objects with varying temperatures are in contact in an isolated system, heat will flow from the hotter object to the cooler one. This flow continues until thermal equilibrium is reached, meaning all parts have the same temperature.

In the given problem, we are looking at several substances exchanging heat: lead, ice, water, and a copper can. The concept of heat flow is dictated by the algebraic sum of heat transfers from each substance. The principle ensures the sum of heat lost by one substance equals the heat gained by others, following the formula:

• For heat lost/gained: \( Q_{ice} + Q_{water} + Q_{copper} = -Q_{lead} \).

This representation mandates that heat lost (negative) by the lead equals the total heat gained by the ice, water, and copper together.

Specific Heat Capacity

Specific heat capacity is an important concept when discussing heat exchange. It is defined as the amount of heat per unit mass required to raise the temperature of a substance by one degree Celsius. It varies with different materials, making it a unique property that plays a key role in heat calculations.

In our problem, each material—lead, water, and copper—has its own specific heat capacity:

• Lead (\(c_{lead}\)): 0.128 J/g°C
• Water (\(c_{water}\)): 4.186 J/g°C
• Copper (\(c_{copper}\)): 0.385 J/g°C

These values are plugged into the heat exchange calculations to determine how much heat each material will absorb or release while changing temperature until equilibrium is reached. As lead cools down, water, ice, and copper heat up to balance the energy flow, driven by these specific heat capacities.

Latent Heat of Fusion

Latent heat of fusion is the heat required to change a substance from solid to liquid at constant temperature, without increasing its temperature. In this exercise, it refers to the amount of heat needed to turn ice into liquid water. This phase change occurs at 0°C and requires energy input.

For ice, the latent heat of fusion is quite high at 334,000 J/kg, and this heat is absorbed as the ice melts. You can describe this process with:

• Ice's heat gain: \( Q_{ice} = m_{ice} \cdot L_f \).

Our exercise calculates this first step before moving onto the subsequent heating of the resulting water. Even though the temperature doesn't change during melting, energy is still spent breaking bonds in the ice, indicating latent energy absorption.

Conservation of Energy

The conservation of energy principle is fundamental to calorimetry problems. It asserts that energy cannot be created or destroyed; it can only be transformed from one form to another. In our context, this means that the heat lost by the hot lead must equal the combined energy gained by the cooler ice, water, and copper.

The initial and final energy levels in this closed system must match. Thus, we use conservation equations to balance the energy flow:

• Heat gained by components (ice, water, copper) = Heat lost by lead.

This ensures that, within the calorimeter, the net energy change is zero, allowing us to solve for unknown quantities, like the final temperature, by balancing energy inputs and outputs accurately.