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Chapter 17: Problem 96

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 \(0.884 \mathrm{~kg} .\) Assuming no heat exchange with the surroundings, what mass of ice was added?

Short Answer

Expert verified
The mass of the ice added from the refrigerator can be calculated from the energy balance equation formed by the above steps. The result should be in Kilograms.

Step by step solution

01

Compute the Heat Absorbed by the Ice to Reach 0 degree Celsius

First, use the equation \(q = m c \Delta T\) to find the heat absorbed by the ice to reach 0 degree Celsius. The specific heat capacity of ice is \(2.09 \, J/g \cdot ^{\circ}C\), the mass of the ice is unknown (let's label it as \(m1\)), and the change in temperature is \(15^{\circ}C\). So, the equation will be: \(q_1 = m1 * 2.09 * 15\). Let's note this as Equation 1.
02

Compute the Heat Absorbed by the Ice to Melt

We know the ice melts completely to water at 0 degree Celsius, so we need to calculate the heat requirement for this process by the formula \(q = mLf\), where \(L_f\) is the heat of fusion, which equals to \(334 J/g\) for ice. So, the equation is \(q_2 = m1 * 334\). This is Equation 2.
03

Compute the Heat Released by the Melting Water

Initially, some ice was present in the water. The heat that melted this ice came from the water itself. \nUse the heat of fusion formula \(q = mLf\) again. We know from the problem description that \(0.884 kg - 0.450 kg = 0.434 kg\) of ice melted. So, the equation is \(q_3 = -0.434 * 10^3 * 334\). The negative sign denotes heat release. Note this as Equation 3.
04

Compute the Heat Released by Water Cooling Down to 0 degree Celsius

Lastly, you have to compute the heat released by the \(1.75 kg\) of water and \(0.434 kg\) of melted ice (which is now water) in cooling down to 0 degree Celsius. Use the heat formula \(q = mc \Delta T\) again. The specific heat of water is \(4.186 J/g \cdot ^{\circ}C\), the mass of the water is \(1.75 kg + 0.434 kg\), the initial temperature of the water is assumed to be room temperature (around \(22^{\circ}C\)). So, the equation is \(q_4 = -(1.75*10^3 + 0.434*10^3) * 4.186 * 22\). The negative sign denotes heat release. Note this as Equation 4.
05

Adding all the Heat Equations

There's no exchange of heat with surroundings, so the heat absorbed by the added ice (Equations 1 and 2) must equal the heat released by the water and melting ice (Equations 3 and 4). Set them equal and solve for \(m1\).
06

Find the mass of added Ice

After calculating \(m1\) in Grams, don't forget to convert it back to Kilograms (since the mass in the problem is given in kilogram).

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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 crucial concept in understanding heat transfer in substances. It defines the amount of heat energy required to change the temperature of a unit mass of a substance by one degree Celsius. For any material, the specific heat capacity is a constant value and varies across different substances.
In our problem, the specific heat capacity of ice is key to determining how much heat is needed to bring more ice from \[ -15^{ ext{掳C}} \] up to 0掳C, where the phase change can happen.

To determine this heat energy (\(q_1\)), we use the equation: \[q = m c \Delta T\]where \(m\) is the mass of the ice, \(c\) is the specific heat capacity (equivalent to \(2.09 \, \text{J/g掳C}\) for ice), and \(\Delta T\) is the change in temperature. This formula helps calculate the energy needed to raise the temperature of ice to its melting point before any phase change occurs. Consequently, specific heat capacity helps us understand and calculate how substances absorb energy when not undergoing phase changes, which is essential when ice transitions from below freezing to melting point.
Heat of Fusion
The heat of fusion plays a pivotal role when materials change state, like ice converting to water. It represents the energy needed to change a unit mass of a substance from solid to liquid without changing its temperature.
In this scenario, after the ice reaches 0掳C, the next step involves melting the ice, for which we apply the heat of fusion. The formula \(q = m L_f\) is used, where \(m\) is the mass of the ice, and \(L_f\) is the heat of fusion. For ice, this value is \(334 \, \text{J/g}\).Essentially, the heat of fusion equates to the energy required per gram to transition ice at 0掳C to water at 0掳C.
  • This energy does not increase the temperature but instead changes the state.
  • Understanding this aids in calculations, predicting how much energy is absorbed without temperature shifts.
Hence, the heat of fusion is paramount in heat transfer calculations, allowing us to figure out how much ice has melted to achieve thermal equilibrium in mixtures.
Thermal Equilibrium
Thermal equilibrium denotes a balanced state where no heat flows between objects, and all parts of the system are at the same temperature. It forms the basis of solving heat transfer problems, such as the presented exercise.
When more ice is added to the bucket, thermal equilibrium ensures that all added ice, initial ice, and water reach the same temperature. This equilibrium implies that the total heat absorbed by the added ice must equal the total heat released by the water and any melting ice.

By understanding thermal equilibrium:
  • We know the system is "insulated," meaning no heat enters or leaves the environment.
  • Calculations focus on internal transfers, balancing absorbed and released heat to solve for the unknown.
In practical terms, achieving thermal equilibrium lets us determine the mass of the ice added by ensuring all energy exchanges within the system are accounted for, leading to an overall balance of energy flows.

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Most popular questions from this chapter

An asteroid with a diameter of \(10 \mathrm{~km}\) and a mass of \(2.60 \times 10^{15} \mathrm{~kg}\) impacts the earth at a speed of \(32.0 \mathrm{~km} / \mathrm{s},\) landing in the Pacific Ocean. If \(1.00 \%\) of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of \(\left.10.0^{\circ} \mathrm{C}\right),\) what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{~kg} .\)

One end of an insulated metal rod is maintained at \(100.0^{\circ} \mathrm{C}\), and the other end is maintained at \(0.00^{\circ} \mathrm{C}\) by an ice-water mixture. The rod is \(60.0 \mathrm{~cm}\) long and has a cross-sectional area of \(1.25 \mathrm{~cm}^{2}\). The heat conducted by the rod melts \(8.50 \mathrm{~g}\) of ice in \(10.0 \mathrm{~min} .\) Find the thermal conductivity \(k\) of the metal.

Convert the following Celsius temperatures to Fahrenheit: (a) \(-62.8^{\circ} \mathrm{C}\), the lowest temperature ever recorded in North America (February \(3,1947,\) Snag, Yukon); (b) \(56.7^{\circ} \mathrm{C},\) the highest temperature ever recorded in the United States (July \(10,1913,\) Death Valley, California); (c) \(31.1^{\circ} \mathrm{C},\) the world's highest average annual temperature (Lugh Ferrandi, Somalia).

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?

There is \(0.050 \mathrm{~kg}\) of an unknown liquid in a plastic container of negligible mass. The liquid has a temperature of \(90.0^{\circ} \mathrm{C}\). To measure the specific heat capacity of the unknown liquid, you add a mass \(m_{w}\) of water that has a temperature of \(0.0^{\circ} \mathrm{C}\) to the liquid and measure the final temperature \(T\) after the system has reached thermal equilibrium. You repeat this measurement for several values of \(m_{\mathrm{w}},\) with the initial temperature of the unknown liquid always equal to \(90.0^{\circ} \mathrm{C}\). The plastic container is insulated, so no heat is exchanged with the surroundings. You plot your data as \(m_{\mathrm{w}}\) versus \(T^{-1}\), the inverse of the final temperature \(T\). Your data points lie close to a straight line that has slope \(2.15 \mathrm{~kg} \cdot \mathrm{C}^{\circ} .\) What does this result give for the value of the specific heat capacity of the unknown liquid?
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