Question

A thirsty nurse cools a 2.00-L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.257 kg and adding 0.120 kg of ice initially at -15.0\(^\circ\)C. If the soft drink and mug are initially at 20.0\(^\circ\)C, what is the final temperature of the system, assuming that no heat is lost?

Short answer

The final temperature of the system is 14.9°C.

Step-by-step solution

Identify Given Information

We have:- Mass of the soft drink (mostly water): 2.00 L \( = 2.00 \text{ kg} \) (assuming density of water is \( 1 \text{ kg/L} \))- Initial temperature of the soft drink: \( 20.0^{\circ}\text{C} \)- Heat capacity of water: \( c_\text{water} = 4.18 \text{ J/g}^\circ\text{C} \)- Mass of the aluminum mug: \( 0.257 \text{ kg} \)- Initial temperature of the aluminum mug: \( 20.0^{\circ}\text{C} \)- Specific heat capacity of aluminum: \( c_\text{Al} = 0.897 \text{ J/g}^\circ\text{C} \)- Mass of the ice: \( 0.120 \text{ kg} \)- Initial temperature of the ice: \( -15.0^{\circ}\text{C} \)- Specific heat capacity of ice: \( c_\text{ice} = 2.09 \text{ J/g}^\circ\text{C} \)- Latent heat of fusion for ice: \( L_f = 334 \text{ J/g} \)

Calculate Heat Exchange by Ice to 0°C

The ice must first warm from \(-15^{\circ}\text{C}\) to \(0^{\circ}\text{C}\). The heat required is given by:\[q_1 = m_\text{ice} \times c_\text{ice} \times \Delta T = 120 \text{ g} \times 2.09 \text{ J/g}^\circ\text{C} \times (0 - (-15)) = 3762 \text{ J}\]

Calculate Heat Required for Melting Ice

The ice at \(0^{\circ}\text{C}\) will melt to become water at \(0^{\circ}\text{C}\). The heat required for this phase change is:\[q_2 = m_\text{ice} \times L_f = 120 \text{ g} \times 334 \text{ J/g} = 40080 \text{ J}\]

Calculate Heat Lost by Water and Mug

The soft drink initially at \(20^{\circ}\text{C}\) will lose heat as it approaches equilibrium temperature \(T_f\):\[q_3 = (m_\text{water} \times c_\text{water} + m_\text{Al} \times c_\text{Al}) \times \Delta T = (2000 \text{ g} \times 4.18 + 257 \text{ g} \times 0.897) \times (20 - T_f)\]Expanding,\[q_3 = (8360 + 230.529) \times (20 - T_f) = 8590.529 \text{ J/}^\circ\text{C} \times (20 - T_f)\]

Equate Heat Gained and Lost

The total heat gained by the ice must equal the heat lost by the water and mug:\[q_1 + q_2 = q_3\]Substitute the values:\[3762 + 40080 = 8590.529 \text{ J/}^\circ\text{C} \times (20 - T_f)\]Solve for \(T_f\):\[43842 = 8590.529 \text{ J/}^\circ\text{C} \times (20 - T_f)\]\[5.104 = 20 - T_f\]\[T_f = 14.9^{\circ}\text{C}\]

Verify the Assumptions

Check to ensure this temperature is feasible and all ice has melted by its energy calculations. The final temperature calculated is above \(0^{\circ}\text{C}\), indicating complete melting occurred, confirming no further adjustments are necessary.

Key concepts

Heat Transfer

Heat transfer is an essential aspect of thermodynamics, responsible for energy exchange between systems at different temperatures. In the discussed scenario, the heat transfer process allows the cold ice and hot soft drink to reach an equilibrium temperature.
The motto of heat transfer is very much like that of energy: "Energy cannot be created or destroyed, only transferred." This transfer aims at establishing thermal equilibrium, where temperatures level out.
There are three modes of heat transfer: conduction, convection, and radiation. In our context, conduction primarily occurs as the mug, ice, and drink share energy until they reach a uniform temperature.
Understanding these concepts helps explain why heat moves from the soft drink and mug, which lose heat, to the ice, which gains heat.

Specific Heat Capacity

Specific heat capacity is a material property that defines how much heat energy is needed to change the temperature of a given mass of a substance by one degree Celsius. It is denoted by the symbol \( c \) and measured in units of J/g°C.
For example, in this exercise, water has a specific heat capacity of 4.18 J/g°C. This means it takes 4.18 Joules of energy to raise the temperature of one gram of water by one degree Celsius.
Similarly, aluminum, with its specific heat capacity of 0.897 J/g°C, requires less energy per gram per degree Celsius change. The differences in specific heat capacities play a crucial part in how quickly or slowly objects warm up or cool down during heat transfer.

• Higher specific heat capacity means a substance stores more heat for a temperature change.
• Metals generally have lower specific heat capacities compared to water, meaning they heat up and cool down more quickly.

In any thermodynamic process involving different substances, it’s crucial to consider these properties to predict how energy will flow through the system.

Latent Heat of Fusion

Latent heat of fusion is the amount of energy required to change a substance from a solid to a liquid at its melting point without changing its temperature. It is crucial during phase changes, such as when ice turns into water.
In this exercise, the ice at 0°C needs 334 J/g to melt into liquid water. Importantly, even though the ice absorbs a large amount of energy, its temperature does not increase until all ice has melted. This unique aspect is what distinguishes latent heat from specific heat.

• Latent heat is only relevant during phase changes.
• No temperature change occurs during the phase change despite heat absorption.

Understanding latent heat of fusion helps explain why melting ice absorbs much more heat than simply warming it to its melting point.

Melting Ice Process

The melting ice process is a fascinating example of heat transfer and phase change working in tandem. Initially, the ice begins at -15°C and must be warmed to 0°C before it can melt. This warming requires a certain amount of energy, calculated using the specific heat capacity of ice.
Once at 0°C, the ice undergoes a phase transformation. To become liquid water, it absorbs latent heat—energy that’s necessary but doesn’t increase temperature. After complete melting, any additional heat changes the temperature of the resulting water.

• Energy is used in two stages: warming to melting point, then phase transition.
• Only after all ice has melted does any temperature rise in liquid water occur.

In thermodynamics problems, calculating the energy needed at each phase is essential to determine how different components will interact and reach equilibrium temperatures.