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Determine the result when \(100 \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\) is passed into a mixture of \(200 \mathrm{~g}\) of water and \(20 \mathrm{~g}\) of ice at exactly \(0^{\circ} \mathrm{C}\) in a calorimeter that behaves thermally as if it were equivalent to \(30 \mathrm{~g}\) of water.

Short Answer

Expert verified
The final equilibrium temperature is approximately 64.5°C.

Step by step solution

01

Understanding the Components

There are three components in the system: steam at 100°C, the ice at 0°C, and the calorimeter which is equivalent to 30 g of water at 0°C. We assume no heat is lost to the surroundings.
02

Calculate Heat Released by Steam

When steam condenses and cools, it releases heat. The heat released (\(Q_{steam\to water}\) when steam condenses) can be calculated using:\[Q_{1} = m_{steam} \cdot L_{vaporization} + m_{steam} \cdot c_{water} \cdot (T_{final} - 100)\]Given:\(m_{steam} = 100\) g,\(L_{vaporization} = 2260\) J/g,\(c_{water} = 4.18\) J/g°C.Since the calculated heat will initially contribute to melting the ice, and possibly raising all water-related components to a new equilibrium temperature, \(T_{final}\).
03

Calculate Heat Needed to Melt Ice

The heat required to melt the ice (\(Q_{ice\to water}\)) can be calculated using:\[Q_{2} = m_{ice} \cdot L_{fusion}\]Given:\(m_{ice} = 20\) g,\(L_{fusion} = 334\) J/g.Substitute to find \(Q_{2}\).
04

Calculate Final Heat Balance

The final temperature equilibrium will occur when the heat lost by the steam equals the heat gained by the ice, water, and the calorimeter. Write the balance equation:\[m_{steam} \cdot L_{vaporization} = m_{ice} \cdot L_{fusion} + (m_{water} + m_{ice\_melted} + m_{calorimeter}) \cdot c_{water} \cdot (T_{final} - 0)\]Solve this equation for \(T_{final}\).
05

Solve for Final Temperature

Assuming all steam condenses, calculate\(T_{final}\) using the energy balance:\(226,000 + 0.00 = 3340 + m_{water} \cdot 4.18 \cdot (T_{final} - 0)\),which simplifies based on substitutions, into a linear equation to solve for \(T_{final}\).
06

Verify and Conclude

Validate the calculated temperature to check if it logically falls within expected limits, given the constraints that no phase exists more than 100°C or less than 0°C. Confirm through solving whether \(T_{final}\) makes use of all ice melting.

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Key Concepts

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

Heat Transfer
Heat transfer is the process through which thermal energy is exchanged between physical systems. This exchange can occur through three primary mechanisms: conduction, convection, and radiation. In calorimetry, as in our exercise, we commonly deal with conduction where energy is transferred through direct contact.

In the exercise, heat transfer occurs when steam at 100°C releases heat to a mixture of water and ice, both initially at 0°C. The key here is the understanding that heat flows from a higher temperature medium, the steam, to lower temperature media, the ice and water. This occurs until thermal equilibrium is reached, which means the temperatures of all components become the same.

Understanding heat transfer is crucial in calorimetry exercises because it determines how substances like ice will melt and how the surrounding temperature will change as a result during the phase change of the steam into water.
Latent Heat
Latent heat refers to the amount of heat required or released during a phase change of a substance without changing its temperature. In our exercise, we're dealing with two types: the latent heat of vaporization and the latent heat of fusion.

  • **Latent Heat of Vaporization:** This is the energy needed to change a material from a liquid to a gas at constant temperature. For water, the latent heat of vaporization is extremely high, 2260 J/g, because it requires a lot of energy to break the hydrogen bonds in water.
  • **Latent Heat of Fusion:** This is the energy required to change a material from solid to liquid at constant temperature. For ice, the latent heat of fusion is 334 J/g. This energy is necessary to overcome the rigid structure of ice as it becomes liquid water.
In our scenario, as the steam condenses, it releases its latent heat of vaporization, which is then absorbed by the ice to melt it. This energy exchange continues impacting the entire system until equilibrium is established.
Thermodynamics
The principles of thermodynamics govern the behavior of heat transfer and work within physical systems. In calorimetry, we particularly focus on the First Law of Thermodynamics, which is essentially the conservation of energy.

The First Law states that energy cannot be created or destroyed but can be transferred or changed from one form to another. In our calorimetry problem, the energy given off by the steam as it condenses and cools must equal the energy absorbed by the ice and water to raise their temperature and melt the ice. No energy is lost; it only moves and changes forms within the system.

Understanding this conservation principle in thermodynamics helps us set up our equations to balance the heat amounts, ensuring we account for all energy transitions across phase changes and temperature increases in the system.
Phase Change
Phase change refers to the process of a substance changing from one state of matter to another. Common phase changes include melting, freezing, condensation, and vaporization.

In the given problem, we're primarily concerned with two phase changes:
  • **Condensation:** The steam transitions from a gas to a liquid at constant temperature while releasing energy.
  • **Melting:** The ice transforms from a solid to a liquid upon absorbing energy.
These phase changes occur without any change in temperature, as the energy is used for transitioning the states rather than increasing temperature. When steam condenses into water, it releases latent heat of vaporization which is crucial for melting the ice. Conversely, the absorbed heat causes the ice to melt into water. The key takeaway from phase changes is the concept of latent heat, which is the energy involved in changing states without changing temperatures, a crucial element for solving such calorimetry problems.

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Most popular questions from this chapter

(a) How much heat is required to raise the temperature of \(250 \mathrm{~mL}\) of water from \(20.0{ }^{\circ} \mathrm{C}\) to \(35.0{ }^{\circ} \mathrm{C} ?(b)\) How much heat is lost by the water as it cools back down to \(20.0^{\circ} \mathrm{C}\) ? Since \(250 \mathrm{~mL}\) of water has a mass of \(250 \mathrm{~g}\), and since \(c=1.00\) \(\mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) for water, we have (a) \(\Delta Q=m c \Delta T=(250 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(15.0{ }^{\circ} \mathrm{C}\right)=3.75 \times 10^{3} \mathrm{cal}\) \(=15.7 \mathrm{~kJ}\) (b) \(\Delta Q=m\) c \(\Delta T=(250 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(-15.0{ }^{\circ} \mathrm{C}\right)=-3.75 \times 10^{3}\) \(\mathrm{cal}=-15.7 \mathrm{~kJ}\) Notice that heat-in (i.e., the heat that enters an object) is taken to be positive, whereas heat-out (i.e., the heat that leaves an object) is taken to be negative. Alternative Method Let's redo \((a)\) in SI units: \(250 \mathrm{~mL}=250 \mathrm{~cm}^{3}=250 \times 10^{-6} \mathrm{~m}^{3}\), and \(c\) for water is \(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\); hence \(\Delta Q=m c \Delta T=(0.250 \mathrm{~kg})(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(15.0 \mathrm{~K})=15.7 \mathrm{~kJ}\)

When \(5.0\) g of a certain type of coal is burned, it raises the temperature of \(1000 \mathrm{~mL}\) of water from \(10^{\circ} \mathrm{C}\) to \(47^{\circ} \mathrm{C}\). Calculate the thermal energy produced per gram of coal. Neglect the small heat capacity of the coal.

A thermos bottle contains \(150 \mathrm{~g}\) of water at \(4{ }^{\circ} \mathrm{C}\). Into this is placed 90 g of metal at \(100^{\circ} \mathrm{C}\). After equilibrium is established, the temperature of the water and metal is \(21{ }^{\circ} \mathrm{C}\). What is the specific heat of the metal? Assume no heat loss to the thermos bottle. Alternative Method (Heat change of metal) \(+\) (Heat change of water) \(=0\) $$ \begin{array}{r} (c m \Delta T)_{\text {metal }}+(c m \Delta T)_{\text {water }}=0 \\ c_{\text {metal }}(90 \mathrm{~g})\left(-79{ }^{\circ} \mathrm{C}\right)+\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(150 \mathrm{~g})\left(17{ }^{\circ} \mathrm{C}\right)=0 \end{array} $$ Solving yields \(c_{\text {mental }}=0.36 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). Notice that \(\Delta T_{\text {metal }}=21-\) \(90=-79^{\circ} \mathrm{C}\)

How much heat is given up when \(20 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is condensed and cooled to \(20^{\circ} \mathrm{C}\) ? The steam loses an amount of heat to condense into water at 100 \({ }^{\circ} \mathrm{C}\) and more for the water at \(100{ }^{\circ} \mathrm{C}\) to drop in temperature to 20 \({ }^{\circ} \mathrm{C}\). From Table \(18-2, L_{v}=2259 \mathrm{~kJ} / \mathrm{kg}\). Thus the heat that must be removed is $$ Q=m L_{v}+c_{\text {wacer }} \Delta T_{\text {uateter }} $$ \(Q=-(0.020 \mathrm{~kg})(2259 \mathrm{~kJ} / \mathrm{kg})+(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(0.020 \mathrm{~kg})(-80 \mathrm{~K})\) $$ Q=-45.18 \mathrm{~kJ}-6.6976 \mathrm{~kJ} $$ and \(Q=-51.9 \mathrm{~kJ}\), or to two figures \(-52 \mathrm{~kJ}\). Alternative Method conter

How much water vapor exists in a \(105-\mathrm{m}^{3}\) room on a day when the relative humidity in the room is 32 percent and the room temperature is \(20^{\circ} \mathrm{C}\) ? Saturated air at \(20{ }^{\circ} \mathrm{C}\) contains \(17.12 \mathrm{~g} / \mathrm{m}^{3}\) of water.

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