Problem 104 - University Physics with Modern Physics

Question

A Styrofoam bucket of negligible mass contains 1.75 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice. More ice, from a refrigerator at \(-15.0^{\circ} \mathrm{C},\) is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.778 \(\mathrm{kg}\) . Assuming no hear exchange with the surroundings, what mass of ice was added?

Step-by-Step Solution

Answer

0.778 kg - 0.450 kg of ice was added.

Step 1: Calculate Heat Required to Melt Existing Ice

We know the existing 0.450 kg of ice must begin melting. We use the heat of fusion for ice, which is 334,000 J/kg. However, in this case, since we end with 0.778 kg of ice, not all the original ice melts.The heat required to melt the existing ice is calculated using:\[ q_1 = m_{ice} \times L_f \]Where:- \( q_1 \) is the heat required to melt part of the ice,- \( m_{ice} \) is the mass of ice melting (since we end up with more ice, \( m_{ice} = 0.450 \text{ kg} - \text{final ice} = 0 \)) because no melting occurs.

Step 2: Calculate Heat Transfer from Added Ice

The added ice comes at \(-15^{\circ}C\). It first needs to be warmed to \(0^{\circ}C\), then may melt. We calculate this with:\[ q_2 = m_{added} \times c_{ice} \times \Delta T \]Where:- \( m_{added} \) is the mass of ice added (unknown),- \( c_{ice} = 2,100 \text{ J/kg°C} \) is the specific heat of ice,- \( \Delta T = 15°C \) is the change in temperature.Since we end with more ice, added ice doesn't entirely melt, so assume the mass of added ice equals the final ice mass minus initial 0.450 kg. Calculate \( q_2 \) once assuming some of it melts or helps cool the water.

Step 3: Consider Heat Exchange with Water

Water at \(0^{\circ}C\) has an energy contribution too. We need to find the heat lost by the water as the system reaches equilibrium:\[ q_3 = m_{water} \times c_{water} \times \Delta T \]Where:- \( m_{water} = 1.75 \text{ kg} \) is the mass of the water,- \( c_{water} = 4,186 \text{ J/kg°C} \) is the specific heat of water,- Let \( \Delta T = 0 \text{ as temperature remains same}.\)No additional heat transfer needed from the water side as ice absorbs it all.

Step 4: Apply Energy Conservation (Combining All Heat Exchanges)

Since the system is closed and reaches thermal equilibrium, we apply energy conservation:\[ q_1 + q_2 = q_3 \]Substituting the known values and solving for \( m_{added} \):-- Initial 0.45 kg stays as solid so not melting for existing but new ice adds.- Final ice mass 0.778 kg is the clue plus initial implies melt some new part.Computations will align roughly with initial ice handling, and main is this equilibrium data to solve added amount to obtain 0.778 kg ultimately.

Step 5: Calculate Mass of Added Ice

Given the condition changes from analytical setup above, new added mass water contribution melting's equilibrium computes to:\[ m_{added} = 0.778 \text{ kg final equilibrium}- 0.450 \text{ kg} \] roughly accounting for losses melting dynamics from \(-15°C\) up. Calculating gives estimated needed to start balancing. Thus ultimately:The calculated parcels simplify to factual delivery.

Key Concepts

Heat of Fusion

The heat of fusion is the amount of energy needed to change a substance from the solid phase to the liquid phase at its melting point. For ice, this is a particularly important concept, as it requires a substantial amount of energy, precisely 334,000 J/kg, to transform from ice to water without changing temperature. This process is crucial in scenarios where ice and water coexist, such as the problem mentioned, as understanding this helps you calculate how much heat energy is needed to melt existing ice or how added ice might interact thermally with the water present. In the example provided, although some ice was initially present, not all of it melted because additional ice was added, resulting in a balance of unmelted ice remaining.

Specific Heat Capacity

Specific heat capacity is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius. This property is essential when determining how substances like water or ice react to heat changes. For water, the specific heat capacity is 4,186 J/kg°C, whereas for ice it is 2,100 J/kg°C. Understanding these values helps in calculating the energy exchange within the system. In the given exercise, you need to know the specific heat capacity of ice to determine how much energy is required to raise its temperature from \(-15^{\circ}C\) to \(0^{\circ}C\). Without fully melting the new ice, this essential step ensures you correctly account for how it adjusts the system’s total energy.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another. It is a key component in reaching thermal equilibrium within a system. In the case of the bucket of water and ice, heat transfer occurs as thermal energy moves from the warmer water to the colder, newly added ice. This affects the overall temperature and phases of the materials present. The mass of the added ice will impact how much energy is transferred. Using the temperature change and specific heat capacity values, you can calculate how much energy will be transferred to bring the newly added ice to the same temperature as the existing materials in the system.

Conservation of Energy

Conservation of energy is a fundamental principle stating that energy cannot be created or destroyed, only transformed from one form to another. In thermal systems, this means the total thermal energy before and after a change remains the same, assuming no heat is lost to the surroundings. In practical terms for the exercise, this principle implies that the heat lost by the water equals the heat gained by the ice. This principle allowed us to calculate the needed energy transfers to re-establish equilibrium. By understanding and applying conservation of energy, you'll determine the mass of ice needed to achieve the final equilibrium with \(0.778\text{ kg}\) of ice remaining, ensuring the entire system's energy is accounted for effectively.