Problem 93 - University Physics with Modern Physics

Question

You have 1.50 kg of water at 28.0\(^\circ\)C in an insulated container of negligible mass. You add 0.600 kg of ice that is initially at -22.0\(^\circ\)C. Assume that no heat exchanges with the surroundings. (a) After thermal equilibrium has been reached, has all of the ice melted? (b) If all of the ice has melted, what is the final temperature of the water in the container? If some ice remains, what is the final temperature of the water in the container, and how much ice remains?

Step-by-Step Solution

Answer

All of the ice doesn't melt; final temperature is 0°C with 0.156 kg of ice remaining.

Step 1: Calculate the energy needed to warm the ice to 0°C

The specific heat capacity of ice is 2,090 J/kg°C. We must first calculate the energy needed to warm 0.600 kg of ice from -22.0°C to 0°C. The formula for this is: \[ q_1 = m imes c imes riangle T = 0.600 imes 2090 imes (0 - (-22)) \] Calculating this gives: \[ q_1 = 0.600 imes 2090 imes 22 = 27,588 \text{ J} \]

Step 2: Calculate the energy needed to melt the ice at 0°C

To melt ice into water, the latent heat of fusion is needed, which is 334,000 J/kg. Using 0.600 kg of ice, we calculate: \[ q_2 = m imes L_f = 0.600 imes 334,000 \] Calculating the energy gives: \[ q_2 = 200,400 \text{ J} \]

Step 3: Calculate the energy available from cooling the water to 0°C

The specific heat of water is 4,186 J/kg°C. To calculate the energy released by 1.50 kg of water cooling from 28.0°C to 0°C, we use the formula: \[ q_3 = m imes c imes riangle T = 1.50 imes 4186 imes (28 - 0) \] The calculation is: \[ q_3 = 1.50 imes 4186 imes 28 = 176,004 \text{ J} \]

Step 4: Compare energy values to determine if all ice melts

Add the energy required to warm the ice to 0°C and melt it, i.e., \( q_1 + q_2 = 27,588 + 200,400 = 227,988 \text{ J} \). Compare this to \( q_3 \), which is 176,004 J. Since 227,988 J (required) is more than 176,004 J (available), not all the ice melts.

Step 5: Determine the final temperature and remaining ice mass

Since not all ice melts, the final temperature is 0°C. The energy available (176,004 J) first raises the temperature of the ice to 0°C (27,588 J) and then goes into melting part of the ice. The energy left to melt the ice: \[ \text{Energy for melting} = 176,004 - 27,588 = 148,416 \text{ J} \] The mass of ice that melts is: \[ m_{\text{melting}} = \frac{148,416}{334,000} \] \[ m_{\text{melting}} = 0.444 \text{ kg} \] Therefore, \( 0.600 - 0.444 = 0.156 \text{ kg} \) of ice remains.

Key Concepts

Latent Heat of Fusion

The concept of latent heat of fusion is fundamental in thermodynamics when dealing with phase transitions from solid to liquid. Latent heat refers to the energy required to change the state of a substance without changing its temperature. For ice turning into water, this process occurs at 0°C, requiring energy input to break the molecular bonds without raising temperature.
The latent heat of fusion for ice is 334,000 J/kg, meaning each kilogram of ice requires 334,000 joules to fully melt into water. In the given problem, 0.600 kg of ice requires:

• Energy to first reach 0°C from -22°C: calculated separately as it's involved with specific heat capacity.
• Then 200,400 J for melting: \( q_2 = 0.600 \times 334,000 = 200,400 \text{ J} \)

Understanding latent heat is crucial to predict whether a substance will undergo a complete phase change under given energy conditions.

Specific Heat Capacity

Specific heat capacity is a material's intrinsic ability to absorb heat energy per unit mass per degree temperature change. In thermodynamics, it is critical for calculating the amount of heat required to raise a substance's temperature.
In this exercise, two specific heat capacities come into play:

• For ice, it is 2,090 J/kg°C, used for calculating the heat required to bring ice from -22°C to 0°C.
• For water, it is 4,186 J/kg°C, applied for determining the energy release as water cools from 28°C to 0°C.

The energy needed to increase a substance's temperature is calculated using the formula \( q = m \times c \times \Delta T \). For example, warming the given 0.600 kg of ice:

• \( q_1 = 0.600 \times 2090 \times 22 = 27,588 \text{ J} \)

Specific heat capacity thus indicates how much heat a substance can store and how it will affect temperature changes in dynamic systems.

Thermal Equilibrium

Thermal equilibrium is reached in a system when no net energy is exchanged between its components, resulting in a uniform temperature throughout. For this to happen, the energy lost by one part must equal the energy gained by the other.
In the problem, thermal equilibrium occurs when the ice and water reach the same temperature — 0°C — because not enough energy is available to melt all the ice.

• The required energy to bring the system to equilibrium conditions is calculated, and excess energy after one component reaches its melting point is used to melt part of the ice.
• Remaining energy is calculated to determine how much ice retains its phase.

Achieving thermal equilibrium in this scenario means checking if energy supplied equals energy required for temperature changes and potential phase changes. Thus, the mixture won't change further after reaching 0°C.

Phase Transition

Phase transition involves the change of a substance from one state of matter to another, such as from solid to liquid or liquid to gas. These transitions occur at specific temperatures for a given substance and require or release latent heat.
In this exercise's context, we examine the melting of ice into water, a solid-to-liquid transition. The process involves:

• Ice absorbing heat to reach its melting point of 0°C.
• Further absorption of latent heat of fusion for the phase transition at constant temperature.

Since there was insufficient energy to melt all the ice fully, the transition paused, resulting in some remaining as a solid once the system achieved thermal equilibrium.
Phase transitions play a key role in various practical applications, like understanding melting processes in climate systems or designing effective thermal storage solutions.