Problem 40 - University Physics with Modern Physics

Question

In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0\(^\circ\)C is mixed with a mass m of water that has an initial temperature of 80.0\(^\circ\)C. No heat is lost to the surroundings. If the final temperature of the system is 28.0\(\circ\)C, what is the mass m of the water that was initially at 80.0\(^\circ\)C?

Step-by-Step Solution

Answer

The mass of the initial water is approximately 0.492 kg.

Step 1: Calculate Heat Required to Warm Ice to 0°C

Calculate the heat needed to raise the ice temperature from -40.0°C to 0°C using the formula for heat transfer: \[ q_1 = m_{ice} \times c_{ice} \times \Delta T \]where \( q_1 \) is the heat, \( m_{ice} = 0.200 \text{ kg} \) is the mass of the ice, \( c_{ice} = 2.09 \text{ J/g°C} \) is the specific heat capacity of ice, and \( \Delta T = 40.0 \text{°C} \) is the temperature change.First, convert 0.200 kg to grams: \( 0.200 \text{ kg} = 200 \text{ g} \).\[ q_1 = 200 \text{ g} \times 2.09 \text{ J/g°C} \times 40.0 \text{°C} = 16720 \text{ J} \]

Step 2: Calculate Heat to Melt Ice at 0°C

Calculate the heat required to melt the ice at 0°C using the formula:\[ q_2 = m_{ice} \times L_f \]where \( q_2 \) is the heat, and \( L_f = 334 \text{ J/g} \) is the latent heat of fusion for ice.\[ q_2 = 200 \text{ g} \times 334 \text{ J/g} = 66800 \text{ J} \]

Step 3: Calculate Heat to Warm Melted Ice to Final Temperature

Calculate the heat required to warm the melted ice (now water) from 0°C to 28°C using:\[ q_3 = m_{water} \times c_{water} \times \Delta T \]where \( c_{water} = 4.18 \text{ J/g°C} \) is the specific heat capacity of water and \( \Delta T = 28.0 \text{°C} \).\[ q_3 = 200 \text{ g} \times 4.18 \text{ J/g°C} \times 28.0 \text{°C} = 23408 \text{ J} \]

Step 4: Total Heat Required by Ice

Add up all the heat quantities:\[ q_{total} = q_1 + q_2 + q_3 \q_{total} = 16720 \text{ J} + 66800 \text{ J} + 23408 \text{ J} = 106928 \text{ J} \]

Step 5: Calculate Heat Lost by Warm Water

The heat lost by the warm water cooling down to 28°C is calculated by:\[ q_{water} = m \times c_{water} \times \Delta T \]where \( \Delta T = 80.0 \text{°C} - 28.0 \text{°C} = 52.0 \text{°C} \).Since the total heat lost by the warm water equals the total heat gained by the ice:\[ 106928 \text{ J} = m \times 4.18 \text{ J/g°C} \times 52.0 \text{°C} \]

Step 6: Solve for Mass of Initial Water

Isolate \( m \) in the equation:\[ m = \frac{106928 \text{ J}}{4.18 \text{ J/g°C} \times 52.0 \text{°C}} \]Calculate:\[ m \approx \frac{106928 \text{ J}}{217.36 \text{ J/g}} \approx 492 \text{ g} \]Convert grams to kilograms:\[ m = 0.492 \text{ kg} \]

Key Concepts

Specific Heat Capacity

Every substance requires a certain amount of heat energy to change its temperature. The specific heat capacity is a measure of how much heat energy is needed to raise 1 gram of a substance by 1°C. It is what gives different substances their resistance to temperature change. When studying heat transfer between substances, this value is critical because it dictates how much heat is absorbed or released during a temperature change.

In our exercise, specific heat capacity plays a central role in calculating the heat exchange process. For ice, the specific heat capacity is 2.09 J/g°C. This means that each gram of ice needs 2.09 joules to raise its temperature by one degree Celsius. Similarly, for water, the value is 4.18 J/g°C, indicating the amount of energy required to heat 1 gram of water by 1°C. These differing values show why water heats up more slowly than some other substances - it requires more energy to do so.

Understanding specific heat capacity is essential:

• It helps determine how much energy is needed to change the temperature of a substance.
• It varies with different substances, e.g., water versus ice.
• It's fundamental in calculating energy exchange in thermal systems.

Latent Heat of Fusion

Latent heat of fusion refers to the energy needed to change a substance from a solid to a liquid at its melting point, without a change in temperature. When ice melts into water, it absorbs heat energy but stays at 0°C until completely converted.

In our problem, to melt the 0.200 kg of ice, we need to calculate using latent heat. The latent heat of fusion for ice is given as 334 J/g. This means that each gram of ice requires 334 joules of energy to transform into water.

Latent heat of fusion is vital because:

• It figures prominently when transitioning between solid and liquid states.
• Without a temperature increase, it requires significant energy input.
• It's crucial for understanding energy dynamics in phase changes.

For example, even after warming the ice to 0°C, we need additional heat energy to change its phase from solid to liquid. This concept makes latent heat significant in thermal calculations, such as predicting how much energy a particular physical transformation will need.

Temperature Change

Temperature change is an everyday experience, but scientifically, it involves understanding the energy transfer necessary to cause changes in degrees. The process by which a substance absorbs or releases heat results in a temperature increase or decrease, which is a critical aspect of heat transfer calculations.

In our exercise, the ice initially at -40°C must be warmed to 0°C. This change necessitates energy input, calculated using the specific heat capacity. Similarly, the water at 80°C is cooling down to 28°C, releasing energy in the process. Thus, temperature change isn't just a number—it explains how and why heat is exchanged.

• It helps determine the direction of energy transfer - from hot to cold substances.
• Plays a critical role in achieving thermal equilibrium, e.g., both substances eventually reaching 28°C.
• Provides a basis for calculating the energy required or released during heating or cooling.

Knowing how to compute and predict temperature changes is significantly practical in designing systems and processes where thermal control is essential.