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You have \(1.50 \mathrm{~kg}\) of water at \(28.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. You add \(0.600 \mathrm{~kg}\) of ice that is initially at \(-22.0^{\circ} \mathrm{C}\). 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?

Short Answer

Expert verified
After thermal equilibrium is reached, all the ice has not melted. The final temperature of the water is 0°C and \(0.414 \mathrm{~kg}\) of ice remains.

Step by step solution

01

Calculate Total Heat Taken in by Ice

First calculate the heat Q1 absorbed by the ice to increase its temperature to 0°C using the formula Q1 = mass * specific heat * change in temperature = \(0.600 \mathrm{~kg}\) * \(2.1 \mathrm{~J/kg°C}\) * \(22 \mathrm{~°C}\) = \(27720 \mathrm{~J}\). Then calculate the heat Q2 absorbed by the ice to melt it into water: Q2 = mass * heat of fusion = \(0.600 \mathrm{~kg}\) * \(334000 \mathrm{~J/kg}\) = \(200400 \mathrm{~J}\). The total heat absorbed by the ice, Q_ice, is then Q1 + Q2.
02

Calculate Total Heat Released by Water

Next calculate the heat Q_released by the water as it cools to 0°C, using the formula Q_released = mass * specific heat * change in temperature = \(1.50 \mathrm{~kg}\) * \(4186 \mathrm{~J/kg°C}\) * \(28 \mathrm{~°C}\) = \(175548 \mathrm{~J}\).
03

Compare Amounts of Heat

Compare the heat Q_ice absorbed by the ice and the heat Q_released by the water. If Q_ice is less than or equal to Q_released, then all of the ice has melted. If Q_ice is greater than Q_released, some ice remains.
04

Final Temperature Calculation

If all of the ice has melted, the final temperature T is 0°C because any extra heat from the water was used to melt the ice. If some ice remains, then all the heat from the water was used to warm and melt part of the ice, so again the final temperature T is 0°C. If the total heat absorbed by the ice is less than the total heat released by the water, then use the formula (Q_released - Q_ice) / (total mass * specific heat of water) to calculate the final temperature of the water above 0°C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
In our everyday lives, heat transfer is a fundamental concept that happens all around us. It is the process by which thermal energy moves from one body or substance to another. This can occur through various methods like conduction, convection, or radiation. In the exercise we're discussing, the heat is transferred from warmer water to colder ice until an equilibrium is reached where no further heat transfer occurs.

When heat moves from the warm water to the cold ice, a balance is sought such that the total heat given away by the water will attempt to equal the total heat absorbed by the ice. Heat always flows from the hotter object to the cooler one until both reach the same temperature, known as thermal equilibrium.

Understanding this process explains why water cools down while ice warms and eventually melts. The fundamental law here is the first law of thermodynamics: energy is conserved in any process, meaning the energy lost by the water will equal the energy gained by the ice.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to change the temperature of a unit mass of the substance by one degree Celsius. Different substances have different specific heat capacities, which means some require more energy to change temperature than others. Generally, water is known to have a high specific heat capacity which makes it an excellent coolant and heat reservoir.

In the exercise, the specific heat capacity of water is given as 4186 J/kg°C, while ice has a lower value of 2100 J/kg°C. This difference is crucial as it underscores how much energy is needed to change the temperature of each substance. When calculating the heat exchange, we use the specific heat capacity in the formula:
  • For warming ice:
    different mass values imes specific heat imes temperature change
  • For cooling water: similar formula but starting with water's values
This entails that the warmer substance (water) will transfer a conscience amount of energy to the cooler one (ice) until they reach thermal equilibrium without exchanging heat with the surrounding environment.
Latent Heat
Latent heat is the energy absorbed or released during a phase change of a substance without changing its temperature. In the problem we're analyzing, the ice, once it reaches 0°C, begins to melt into water. This transition requires energy called the latent heat of fusion.

The latent heat of fusion is the energy necessary to transform solid ice to liquid water at the same temperature. In the exercise, this amount is specifically 334,000 J/kg for ice. This is crucial because even though the temperature remains constant at the melting point, energy is continually absorbed to facilitate the phase change.

During this process:
  • Ice absorbs heat first to warm up to 0°C.
  • Then, it uses the latent heat to change state from solid to liquid.
This heat balance is critical since it dictates whether all of the ice melts and impacts the final equilibrium temperature of the mixture. In calculations, it's important to account for the latent heat to properly determine the final state of the system.

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Most popular questions from this chapter

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