Problem 44 - University Physics with Modern Physics

Question

In a container of negligible mass, 0.200 \(\mathrm{kg}\) of ice at an initial temperature of \(-40.0^{\circ} \mathrm{C}\) is mixed with a mass \(m\) of water that has an initial temperature of \(80.0^{\circ} \mathrm{C}\) . No heat is lost to the surroundings. If the final temperature of the system is \(20.0^{\circ} \mathrm{C},\) what is the mass \(m\) of the water that was initially at \(80.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The mass of the water initially at 80°C is 0.400 kg.

Step 1: Understanding the Heat Transfer

Since no heat is lost to the surroundings, the heat gained by the ice must be equal to the heat lost by the water. We calculate the heat required to raise ice from \(-40.0^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\), melt it, and then raise the resulting water to \(20.0^{\circ}\mathrm{C}\). We equate this to the heat lost by the water as it cools from \(80.0^{\circ} \mathrm{C}\) to \(20.0^{\circ}\).

Step 2: Heat Required to Warm Ice to 0°C and Melt it

Calculate the heat to warm ice: \[ q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T \]where \( m_{\text{ice}} = 0.200\, \mathrm{kg}, \ c_{\text{ice}} = 2.1\, \mathrm{J/g·^{\circ}C}, \ \Delta T = 40^{\circ} \mathrm{C}\).Convert mass to grams: \( 0.200 \times 1000 = 200 \mathrm{g}\).So, \[ q_1 = 200 \cdot 2.1 \cdot 40 = 16800 \, \mathrm{J} \]Then, calculate the heat to melt ice:\[ q_2 = m_{\text{ice}} \cdot L_f \]where \( L_f = 334\, \mathrm{J/g}\).\[ q_2 = 200 \cdot 334 = 66800 \, \mathrm{J} \]

Step 3: Heat Required to Raise Melted Water to 20°C

Calculate the heat for raising the temperature of water from \(0^{\circ} \mathrm{C}\) (after melting) to \(20^{\circ} \mathrm{C}\):\[ q_3 = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T \]where \(c_{\text{water}} = 4.18\, \mathrm{J/g·^{\circ}C}\) and \(\Delta T = 20^{\circ} \mathrm{C}\).\[ q_3 = 200 \cdot 4.18 \cdot 20 = 16720 \, \mathrm{J} \]

Step 4: Total Heat Absorbed by the Ice

Sum up all the heats calculated:\[ q_{\text{total}} = q_1 + q_2 + q_3 = 16800 + 66800 + 16720 = 100320 \, \mathrm{J} \]

Step 5: Heat Lost by the Water Initially at 80°C

Let's calculate the heat lost by the warmer water:\[ q_{\text{water}} = m \cdot c_{\text{water}} \cdot \Delta T \]where \(\Delta T = 60^{\circ} \mathrm{C}\) (from \(80^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\)).\[ q_{\text{water}} = m \cdot 4.18 \cdot 60 = 250.8m \, \mathrm{J} \]

Step 6: Set the Heat Gained by Ice Equal to Heat Lost by Water

Since all the heat gained by the ice comes from the water:\[ q_{\text{total}} = q_{\text{water}} \]\[ 100320 = 250.8m \]Solve for \( m \):\[ m = \frac{100320}{250.8} = 400.0 \, \mathrm{g} \]Convert to kg: \[ 400 \, \mathrm{g} = 0.400 \, \mathrm{kg} \].

Key Concepts

Thermodynamics

Thermodynamics is the study of energy and its transformations. It focuses on understanding how energy moves and changes within a given system. One important aspect of thermodynamics is heat transfer. Heat transfer refers to the movement of thermal energy from one object or substance to another, based on temperature differences.
In the given exercise, heat transfer plays a central role. The heat lost by the warmer water is transferred to the ice, causing it to eventually reach an equilibrium temperature with the water at 20°C. This performance demonstrates the First Law of Thermodynamics, also known as the principle of energy conservation. It states that, within an isolated system, energy can neither be created nor destroyed; it can only be transferred or transformed.
Every problem involving thermodynamics requires analyzing the transfer of energy between substances and ensuring that this energy conforms to the laws of thermodynamics.

Specific Heat Capacity

Specific heat capacity refers to the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. Different materials absorb and release heat at different rates, and their specific heat capacities greatly influence these rates.
In this exercise, we used specific heat capacities for ice and water to calculate the thermal energy changes. The specific heat capacity of ice is lower than that of water, which means it requires less energy to raise the temperature of ice compared to an equivalent mass of water over the same temperature range:

• Specific heat capacity of ice: 2.1 J/g°C
• Specific heat capacity of water: 4.18 J/g°C

The high heat capacity of water allows it to store considerable energy, explaining why water is effective in regulating Earth's climate and providing natural thermal stability.

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. These transitions require or release heat without changing the temperature of the substance during the process.
In the exercise, we observed a phase change as ice melted to become water. The energy required for this phase change is known as the latent heat of fusion. This value represents the amount of heat needed to convert ice at 0°C to water at the same temperature:

• Latent heat of fusion ( L_f ): 334 J/g

During phase change, energy goes into breaking or forming intermolecular bonds, rather than increasing the substance's kinetic energy, which is why the temperature remains constant. Understanding phase changes is critical in thermodynamics, as they occur widely in nature and technological processes.

Energy Conservation

Energy conservation is a core concept in physics, which states that the total energy of an isolated system remains constant, although energy can transform from one form to another. This principle underpins the First Law of Thermodynamics.
In the problem, no heat is lost to the surroundings, highlighting energy conservation. The energy absorbed by the ice and cold water exactly equals the energy lost by the hot water. This approach lets us set up the equation:\( q_{\text{total}} = q_{\text{water}} \)and find the mass of the warmer water.
Energy conservation principles are fundamental to solving any heat transfer problems in thermodynamics, as they ensure that all energy exchanges are accounted for in closed systems.