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How much water remains unfrozen after \(50.2 \mathrm{~kJ}\) is transferred as heat from \(260 \mathrm{~g}\) of liquid water initially at its freezing point?

Short Answer

Expert verified
Approximately 109.7 g of water remains unfrozen.

Step by step solution

01

Identify the physical properties

The amount of water is given as \(260 \mathrm{~g}\). The heat transferred is \(50.2 \mathrm{~kJ}\). It is necessary to use the latent heat of fusion of water, which is \(334 \mathrm{~J/g}\) (equivalent to \(334000 \mathrm{~J/kg}\)).
02

Convert heat into compatible units

Convert the heat transferred from kilojoules to joules. This can be done by multiplying by \(1000\):\[50.2 \mathrm{~kJ} = 50.2 \times 1000 \mathrm{~J} = 50200 \mathrm{~J}\]
03

Calculate the mass of water that can be frozen

Use the formula relating the heat transferred to the mass of water frozen: \[Q = m \cdot L_f\]where \(Q\) is the heat transferred, \(m\) is the mass of water that freezes, and \(L_f\) is the latent heat of fusion. Rearrange and calculate \(m\):\[m = \frac{Q}{L_f} = \frac{50200 \mathrm{~J}}{334 \mathrm{~J/g}} \approx 150.3 \mathrm{~g}\]
04

Calculate the unfrozen water mass

Subtract the mass of water that freezes from the initial mass to find the unfrozen water:\[260 \mathrm{~g} - 150.3 \mathrm{~g} = 109.7 \mathrm{~g}\]

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Key Concepts

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

Understanding Heat Transfer
Heat transfer is the process by which thermal energy moves from one substance to another. This can happen through conduction, convection, or radiation. When studying heat transfer in the context of melting or freezing, like in our water example, we focus on the conduction part.
It involves moving heat energy from the warmer molecules to the colder ones, effectively changing their temperature or state. In our exercise, we transferred heat energy to the water to see how much of it would freeze.
  • Units of heat are typically Joules (J) or kilojoules (kJ).
  • The process obeys the law of conservation of energy, where energy transferred equals the change in the energy within the substance.
By understanding heat transfer, we can calculate how many grams of water will undergo a phase change when a specific amount of energy is input or removed.
Phase Change and Latent Heat of Fusion
Phase change involves a substance transitioning from one state of matter to another, such as from liquid to solid or vice versa. At the molecular level, this means breaking or forming intermolecular forces without changing temperature. For water freezing, this process involves the latent heat of fusion. The latent heat of fusion is the amount of heat needed to change a certain mass of a substance from solid to liquid without temperature change.
  • For water, this value is 334 J/g. This measurement tells us how much energy is required to freeze or melt each gram of water.
  • During a phase change, temperature remains constant. This is why we still have some unfrozen water even though the energy is removed.
Understanding the energy associated with phase changes helps solve exercises involving heating, freezing, or melting substances.
Mass Calculation in the Context of Heat Transfer
Mass calculation is critical when determining how much of a substance undergoes a phase change. In our problem, it dictates the amount of water freezing upon the withdrawal of heat energy. We must consider both the initial mass and the final resulting mass post-energy transfer.To calculate the freezing water mass, we use the formula: \[ Q = m \cdot L_f \] Where \( Q \) is the heat energy transferred, \( m \) is the mass of freezing water, and \( L_f \) is the latent heat of fusion. By rearranging the formula, we can estimate the frozen mass as \( m = \frac{Q}{L_f} \). Calculating correctly ensures knowing precisely how much water changes state when subjected to a certain amount of heat transfer. This step is crucial for confirming that our calculations of unfrozen water mass post-heat removal match the real-world scenario.
Energy Conversion and Unit Consistency
Energy conversion and unit consistency are important for performing accurate calculations in thermodynamics. Since energy can be measured in different units, converting them is often necessary to maintain consistency and clarity. In our example,
  • Initially, energy is given in kilojoules (kJ), which is common for larger heat values calculations.
  • For compatibility with mass and latent heat of fusion, converting to joules (J) is essential. This step ensures that the formula for heat transfer is dimensionally consistent: 1 kJ = 1000 J.
Maintaining consistent units allows us to apply formulas correctly and systematically solve problems involving energy exchanges, ensuring our calculated values are valid and reliable.

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Most popular questions from this chapter

A solid cylinder of radius \(r_{1}=2.5 \mathrm{~cm}\), length \(h_{1}=5.0 \mathrm{~cm}\), emissivity \(0.85\), and temperature \(30^{\circ} \mathrm{C}\) is suspended in an environment of temperature \(50^{\circ} \mathrm{C}\). (a) What is the cylinder's net thermal radiation transfer rate \(P_{1} ?(\mathrm{~b})\) If the cylinder is stretched until its radius is \(r_{2}=0.50 \mathrm{~cm}\), its net thermal radiation transfer rate becomes \(P_{2}\). What is the ratio \(P_{2} / P_{1}\) ?

Suppose that you intercept \(5.0 \times 10^{-3}\) of the energy radiated by a hot sphere that has a radius of \(0.020 \mathrm{~m}\), an emissivity of \(0.80\), and a surface temperature of \(500 \mathrm{~K}\). How much energy do you intercept in \(2.0 \mathrm{~min} ?\)

A cold beverage can be kept cold even on a warm day if it is slipped into a porous ceramic container that has been soaked in water. Assume that energy lost to evaporation matches the net energy gained via the radiation exchange through the top and side surfaces. The container and beverage have temperature \(T=15^{\circ} \mathrm{C}\), the environment has temperature \(T_{\mathrm{env}}=32^{\circ} \mathrm{C}\), and the container is a cylinder with radius \(r=2.2 \mathrm{~cm}\) and height \(10 \mathrm{~cm}\). Approximate the emissivity as \(\varepsilon=1\), and neglect other energy exchanges. At what rate \(d m / d t\) is the container losing water mass?

A \(1700 \mathrm{~kg}\) Buick moving at \(83 \mathrm{~km} / \mathrm{h}\) brakes to a stop, at uniform deceleration and without skidding, over a distance of \(93 \mathrm{~m}\). At what average rate is mechanical energy transferred to thermal energy in the brake system?

In the extrusion of cold chocolate from a tube, work is done on the chocolate by the pressure applied by a ram forcing the chocolate through the tube. The work per unit mass of extruded chocolate is equal to \(p / \rho\), where \(p\) is the difference between the applied pressure and the pressure where the chocolate emerges from the tube, and \(\rho\) is the density of the chocolate. Rather than increasing the temperature of the chocolate, this work melts cocoa fats in the chocolate. These fats have a heat of fusion of \(150 \mathrm{~kJ} / \mathrm{kg}\). Assume that all of the work goes into that melting and that these fats make up \(30 \%\) of the chocolate's mass. What percentage of the fats melt during the extrusion if \(p=5.5\) MPa and \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}\) ?

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