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Chapter 11: Problem 72

ecp An ice-cube tray is filled with \(75.0 \mathrm{~g}\) of water. After the filled tray reaches an equilibrium temperature \(20.0^{\circ} \mathrm{C}\), it is placed in a freezer set at \(-8.00^{\circ} \mathrm{C}\) to make ice cubes. (a) Describe the processes that occur as energy is being removed from the water to make ice, (b) Calculate the energy that must be removed from the water to make ice cubes at \(-8.00^{\circ} \mathrm{C}\).

Short Answer

Expert verified
To find the total energy removal, we therefore add the energy changes - calculated previously in steps 1, 2, and 3. Adding these gives the total energy that must be removed to change the ice-cube tray filled with \(75.0 \mathrm{~g}\) of water from a temperature of \(20.0^{\circ} \mathrm{C}\) to \(-8.00^{\circ} \mathrm{C}\).

Step by step solution

01

Calculating energy loss during water cooling

We need to calculate the amount of energy loss when water cools from \(20.0^{\circ} \mathrm{C}\) (or 293.15 K, converting to Kelvin) to \(0^{\circ} \mathrm{C}\) (or 273.15 K). The formula for energy loss or gain, \(Q\), is \(Q = mc\Delta T\), where \(m\) is the mass (75 g), \(c\) is the specific heat capacity of water (4.18 J/g.K) and \(\Delta T\) is the change in temperature, \(293.15K - 273.15K = 20K\).
02

Calculating energy loss during phase change

Next, we find the energy loss when water changes phase from liquid to ice at \(0^{\circ} \mathrm{C}\). Here, the formula for energy loss or gain, \(Q\), is \(Q = mL\), where \(m\) is the mass (75 g) and \(L\) is the specific latent heat of fusion for water (334 J/g).
03

Calculating energy loss for ice cooling

Lastly, we calculate the energy loss when ice cools from \(0^{\circ} \mathrm{C}\) to \(-8.00^{\circ} \mathrm{C}\) (or -8 K). Again, we use the formula \(Q = mc\Delta T\) where \(m\) is the mass (75 g), \(c\) is the specific heat capacity of ice (2.09 J/g.K) and \(\Delta T\) is the change in temperature, \(-8 K\).
04

Total energy removal

The total energy that must be removed from the water to make ice cubes at \(-8.00^{\circ} \mathrm{C}\) is given by the sum of the energy changes from Steps 1, 2, and 3. To get the final answer, the individual energy changes are computed using the equations from the steps and then added together.

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Key Concepts

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

Specific Heat Capacity
The specific heat capacity is a property that tells us how much heat energy is needed to change the temperature of a substance. For water, the specific heat capacity is relatively high, which means it takes more energy to change its temperature.
This is why water is used as a coolant in many situations, as it can absorb a lot of heat before it gets warmer.
  • The specific heat capacity of water is 4.18 J/g.K.
  • The formula to calculate energy change is: \[Q = mc\Delta T\]where \(Q\) is the energy in joules, \(m\) is the mass in grams, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature in Kelvin.
So, if you have 75 grams of water cooling from 20掳C to 0掳C, energy is removed based on this capacity. The concept is simple: more mass or a greater temperature change requires more energy removal.
Latent Heat of Fusion
Latent heat of fusion is the heat energy required to change a substance from solid to liquid or vice versa without changing the temperature. It's crucial when we are talking about phase transitions, such as water turning into ice.
  • For water, the latent heat of fusion is 334 J/g.
  • The formula used to calculate this energy change is: \[Q = mL\]where \(Q\) is the energy in joules, \(m\) is the mass in grams, and \(L\) is the latent heat of fusion.
When water reaches 0掳C, it does not immediately become ice, even though it's cold. It remains at 0掳C until enough heat energy is removed to complete the phase change into ice.
Phase Change
A phase change is when a substance moves from one state of matter to another. This usually involves a transfer of energy.
Water turning into ice is an example of phase change where liquid turns into a solid. Even though the temperature stays constant during the phase change from water to ice, energy is still involved in breaking or forming the bonds between molecules.
  • During this transition, the temperature remains constant at 0掳C until all the water becomes ice.
  • After enough energy is removed, the water molecules slow down enough to arrange into a crystal structure, becoming ice.
Phase changes are pivotal in understanding energy transfer because they illustrate how energy is consumed without a change in temperature.
Temperature Change
Temperature change refers to the difference between the initial and final temperatures of a substance as it either gains or loses heat.
This concept underpins calculations of energy transfer during different stages of heating or cooling.
  • The formula still applies: \[Q = mc\Delta T\]where \(\Delta T\) is the temperature difference.
  • In our case, water cools initially from 20掳C to 0掳C, which is a significant change that involves removing energy, and then further to -8掳C as ice.
Since cold temperatures require removing energy, understanding temperature change is key to calculating total energy removal needed to cool substances.

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Most popular questions from this chapter

8 A \(60.0\) -kg runner expends \(300 \mathrm{~W}\) of power while running a marathon. Assuming \(10.0 \%\) of the energy is deliyered to the muscle tissue and that the excess energy is removed from the body primarily by sweating, determine the volume of bodily fluid (assume it is water) lost per hour. (At \(37,0^{\circ} \mathrm{C}\), the latent heat of vaporization of water is \(\left.2.41 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\right)\)

A. \(100-\mathrm{g}\) cube of ice at \(0^{\circ} \mathrm{C}\) is dropped into \(1.0 \mathrm{~kg}\) of water that was originally at \(80^{\circ} \mathrm{C}\). What is the final temperature of the water after the ice has melted?

Calculate the temperature at which a tungsten filament that has an emissivity of \(0.90\) and a surface area of \(2.5 \times 10^{-5} \mathrm{~m}^{2}\) will radiate energy at the rate of \(25 \mathrm{~W}\) in a room where the temperature is \(22^{\circ} \mathrm{C}\).

B. For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for \(24 \mathrm{~h}\) at \(37^{\circ} \mathrm{C}\). A standard constant-temperature bath with electric heating and thermostatic control is not suitable in developing nations without continuously operating electric power lines, Peace Corps volunteer and MIT engineenAmy Smith invented a low-cost, low-maintenance incubator to fill the need. The device consists of a foaminsulated box containing several packets of a waxy material that melts at \(37.0^{\circ} \mathrm{C}\), interspersed among tubes, dishes, or bottles containing the test samples and growth medium (food for bacteria). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then it is put into the box to keep the test samples warm as it solidifies. The heat of fusion of the phasechange material is \(205 \mathrm{~kJ} / \mathrm{kg}\). Model the insulation as a panel with surface area \(0.490 \mathrm{~m}^{2}\), thickness \(9.50 \mathrm{~cm}\), and conductivity \(0.0120 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\). Assume the exterior temperature is \(23.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\) and \(16.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\). (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Explain why your calculation can be done without knowing the mass of the test samples or of the insulation.

A \(75-g\) ice cube at \(0^{\circ} \mathrm{C}\) is placed in \(825 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\). What is the final temperature of the mixture?
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