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Chapter 17: Problem 102

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 0.868 \(\mathrm{kg} .\) Assuming no heat exchange with the surroundings, what mass of ice was added?

Short Answer

Expert verified
About 0.382 kg of ice was added.

Step by step solution

01

Understand the Energy Balance

The system reaches thermal equilibrium when the heat gained by ice from the refrigerator equals the heat lost by the initial water and ice in the bucket. We'll use this principle to set up our equations.
02

Calculate Heat Required to Warm Added Ice to 0掳C

The ice added is initially at \(-15掳C\). To warm it to 0掳C, we use the specific heat capacity of ice (\(c_{ice} = 2.1 \, \mathrm{J/g掳C}\)).Let the mass of ice added be \(m_{added}\). The heat required is: \[ q_1 = m_{added} \times c_{ice} \times (0 - (-15)) \] \[ q_1 = m_{added} \times 2.1 \times 15 \].
03

Calculate Heat Required to Melt the Added Ice

Once the added ice is at 0掳C, it needs heat to melt. The heat required to melt ice is determined by its latent heat of fusion (\(L_f = 334 \, \mathrm{J/g}\)). The heat required is:\[ q_2 = m_{added} \times L_f \]\[ q_2 = m_{added} \times 334 \].
04

Calculate Heat Lost by Initial Water and Ice

Initially, the bucket contains 1.75 kg of water and 0.450 kg of ice at 0掳C. The total mass of ice at thermal equilibrium is 0.868 kg, so 0.868 - 0.450 = 0.418 kg of the water must have turned into ice. The heat lost by the water is:\[ q_{lost} = m_{water\,to\,ice} \times L_f \]where \(m_{water\,to\,ice} = 0.418 \, \mathrm{kg} \), so\[ q_{lost} = 0.418 \times 334 \times 1000 \] (converting kg to g for calculations).
05

Set Up the Energy Balance Equation

We have the expressions for all heats: \[ q_1 + q_2 = q_{lost} \]Substitute the expressions:\[ m_{added} \times (2.1 \times 15 + 334) = 0.418 \times 334 \times 1000 \].
06

Solve for the Mass of Added Ice

Simplify and solve the equation:\[ m_{added} \times (31.5 + 334) = 139612 \]\[ m_{added} \times 365.5 = 139612 \]\[ m_{added} = \frac{139612}{365.5} \approx 381.99 \, \mathrm{g} \].Convert to kg to stay consistent with units: \( m_{added} \approx 0.382 \, \mathrm{kg} \).

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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 key concept in understanding how materials absorb and release heat. It tells us how much heat energy is required to change the temperature of a substance. In this exercise, we focus on the specific heat capacity of ice, which is 2.1 J/g掳C.

This value helps us calculate how much energy the added ice requires to rise from its starting temperature of -15掳C to 0掳C, before any melting occurs.

The formula used is:
  • \( q_1 = m_{added} \times c_{ice} \times (T_{final} - T_{initial}) \)
where \( T_{final} \) is 0掳C, and \( T_{initial} \) is -15掳C.

Understanding this principle allows us to calculate the energy exchange needed to bring the ice to a melt-ready state.
Latent Heat of Fusion
Latent heat of fusion is the energy required to change a substance from solid to liquid without changing its temperature. For ice, this value is 334 J/g.

In the problem, once the added ice reaches 0掳C, it requires additional heat to melt. This is where the latent heat of fusion comes into play. The formula used here is:
  • \( q_2 = m_{added} \times L_f \)


This illustrates the heat necessary for phase change, indicating that even at 0掳C, energy is still required to alter the state of the substance, a concept crucial to understanding the total heat required in this exercise.
Energy Balance
The concept of energy balance is vital for solving this type of thermal equilibrium problem. It involves ensuring the total energy input equals the total energy output, keeping the system stable.

When thermal equilibrium is achieved, the heat gained by the system equals the heat lost. For our exercise, the heat from the added ice warming and melting must balance out with the heat lost by the water turning into more ice. This is expressed in the equation:
  • \( q_1 + q_2 = q_{lost} \)


Using this balance, we deduced how much ice was added to the system, showcasing how crucial energy balance is in applying thermodynamics principles effectively.
Heat Exchange
Heat exchange is a fundamental process where energy is transferred between substances resulting in temperature change or phase change.

In this scenario, the ice and water in the bucket undergo heat exchange to reach thermal equilibrium. The ice absorbs heat from the initial mixture of water and ice in the bucket. This transfer of heat affects both entities, warming the ice and cooling the water until equilibrium is reached.

The situation describes two main types of energy transfer:
  • Warming the ice to 0掳C 鈥� involving sensible heat change, calculated by specific heat capacity.
  • Melting the ice at 0掳C 鈥� involving latent heat, calculated via the latent heat of fusion.


Understanding these exchanges helps clarify the steps we took to determine how much additional ice was added to the system.

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Most popular questions from this chapter

The Sizes of Stars. The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume \(e=1\) for these surfaces. Find the radii of the following stars (assumed to be spherical): (a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of \(2.7 \times 10^{32} \mathrm{W}\) and has surface temperature \(11,000 \mathrm{K}\) ; (b) Procyon \(\mathrm{B}\) (visible only using a telescope), which radiates energy at a rate of \(2.1 \times 10^{23} \mathrm{W}\) and has surface temperature \(10,000 \mathrm{K}\) (c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon \(\mathrm{B}\) is an example of a white dwarf star.)

In a container of negligible mass, 0.200 \(\mathrm{kg}\) of ice at an initial temperature of \(-40.0^{\circ} \mathrm{C}\) is mixed with a mass \(m\) of water that has an initial temperature of \(80.0^{\circ} \mathrm{C}\) . No heat is lost to the surroundings. If the final temperature of the system is \(20.0^{\circ} \mathrm{C},\) what is the mass \(m\) of the water that was initially at \(80.0^{\circ} \mathrm{C} ?\)

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C}\) . By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C}\) ?

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine \(\left(0^{\circ} \mathrm{R}\right)\) However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) . The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.
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