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Chapter 17: Problem 59

A 4.00 -kg silver ingot is taken from a furnace, where its temperature is \(750.0^{\circ} \mathrm{C},\) and placed on a large block of ice at \(0.0^{\circ} \mathrm{C}\) . Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

Short Answer

Expert verified
2.1 kg of ice is melted.

Step by step solution

01

Identify Given Values

We have a 4.00-kg silver ingot with an initial temperature of \(750.0^\circ C\). The specific heat capacity of silver is \(c = 234 \text{ J/kg}^\circ C\). The ingot is placed on ice at \(0.0 ^\circ C\).
02

Calculate Heat Released

We calculate the heat \(Q\) released when the silver is cooled from \(750.0^\circ C\) to \(0.0^\circ C\) using the formula: \[ Q = mc\Delta T, \] where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. Plugging in the values: \[Q = 4.00 \text{ kg} \times 234 \text{ J/kg}^\circ C \times (750.0 - 0.0)^\circ C = 702,000 \text{ J}.\]
03

Determine Ice Melting

The heat required to melt ice is given by the formula \[ Q = m_i L_f, \]where \(m_i\) is the mass of the ice melted and \(L_f = 334,000 \text{ J/kg}\) is the latent heat of fusion for ice. Solve for \(m_i\):\[702,000 \text{ J} = m_i \times 334,000 \text{ J/kg}. \\Rightarrow m_i = \frac{702,000}{334,000} = 2.1 \text{ kg}.\]
04

Conclude

The mass of the ice melted is \(2.1\) kg. Since all the heat from the silver goes into melting the ice, this is the final amount of ice melted.

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Key Concepts

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

Understanding Specific Heat Capacity
Specific heat capacity is a concept in thermodynamics that refers to the amount of heat energy required to change the temperature of a unit mass of a substance by one degree Celsius. This property is crucial because it determines how a material will respond to heat, which is especially important in scenarios like the one described in the exercise with a silver ingot and ice.

For example, silver has a specific heat capacity of 234 J/kg掳C, meaning it takes 234 joules of energy to raise the temperature of one kilogram of silver by one degree Celsius. In the exercise, knowing this value allows us to calculate how much heat the 4 kg silver ingot loses as it cools from 750掳C to 0掳C.

Specific heat capacity can vary greatly between different materials, which is why different substances heat up and cool down at different rates. Understanding this property helps in predicting these temperature changes in different contexts.
Concept of Heat Transfer
Heat transfer is a fundamental principle of thermodynamics that describes how thermal energy moves from a hotter object to a cooler one until thermal equilibrium is reached. This concept is essential in this exercise as it describes the process of the silver ingot transferring heat to the ice.

In our scenario, the heat transfer occurs from the hot silver ingot to the cold ice block. We calculate the amount of heat released using the formula: \[ Q = mc\Delta T \] where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.

Heat transfer occurs through different modes such as conduction, convection, and radiation. In the case of the silver-ice interaction, it primarily occurs through conduction, as the two materials are in direct contact, allowing the heat to flow from the silver to the ice.
Latent Heat of Fusion: Melting Ice
Latent heat of fusion is a term describing the amount of energy required to change a substance from a solid to a liquid at its melting point, without changing its temperature. This concept is crucial in understanding how much ice melts when the silver ingot's heat is transferred.

For ice, the latent heat of fusion is 334,000 J/kg. This means that 334,000 joules are needed to completely melt one kilogram of ice at 0掳C. In the exercise, all of the heat lost by the silver is used to melt the ice.
  • The energy needed to melt 2.1 kg of ice equals the energy released by the ingot, which we've calculated as 702,000 J.
  • By rearranging the fusion formula \( Q = m_i L_f \), you solve for the mass of the ice, \( m_i \), confirming how much ice melts.
Understanding latent heat helps explain energy changes during phase transitions, highlighting why temperature remains constant during melting. This accounts for how the silver ingot can supply sufficient energy to change the ice from its solid state to a liquid state.

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Most popular questions from this chapter

A nail driven into a board increases in temperature. If we assume that 60\(\%\) of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 \(\mathrm{m} / \mathrm{s}\) is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an \(8.00-\mathrm{g}\) aluminum nail after it is struck ten times?

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{K}) ;(\mathrm{c})\) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{K}\right)\).

Why Do the Seasons Lag? In the northern hemisphere, June 21 (the summer solstice) is both the longest day of the year and the day on which the sun's rays strike the earth most vertically, hence delivering the greatest amount of heat to the surface. Yet the hottest summer weather usually occurs about a month or so later. Let us see why this is the case. Because of the large specific heat of water, the oceans are slower to warm up than the land (and also slower to cool off in winter). In addition to perusing pertinent information in the tables included in this book, it is useful to know that approximately two-thirds of the earth's surface is ocean composed of salt water having a specific heat of 3890 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that the oceans, on the average, are 4000 m deep. Typically, an average of 1050 \(\mathrm{W} / \mathrm{m}^{2}\) of solar energy falls on the earth's surface, and the oceans absorb essentially all of the light that strikes them. However, most of that light is absorbed in the upper 100 \(\mathrm{m}\) of the surface. Depths below that do not change temperature seasonally. Assume that the sunlight falls on the surface for only 12 hours per day and that the ocean retains all the heat it absorbs. What will be the rise in temperature of the upper 100 \(\mathrm{m}\) of the oceans during the month following the summer solstice? Does this seem to be large enough to be perceptible?

Spacecraft Reentry. A spacecraft made of aluminum circles the earth at a speed of 7700 \(\mathrm{m} / \mathrm{s} .\) (a) Find the ratio of its kinetic energy to the energy required to raise its temperature from \(0^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\) . (The melting point of aluminum is \(660^{\circ} \mathrm{C}\) . Assume a constant specific heat of 910 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) (b) Discuss the bearing of your answer on the problem of the reentry of a manned space vehicle into the earth's atmosphere.

A laboratory technician drops a \(0.0850-\mathrm{kg}\) sample of unknown solid material, at a temperature of \(100.0^{\circ} \mathrm{C},\) into a calorimeter. The calorimeter can, initially at \(19.0^{\circ} \mathrm{C},\) is made of 0.150 \(\mathrm{kg}\) of copper and contains 0.200 \(\mathrm{kg}\) of water. The final temperature of the calorimeter can and contents is \(26.1^{\circ} \mathrm{C}\) . Compute the specific heat of the sample.
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