/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 An ice-cube tray of negligible m... [FREE SOLUTION] | 91Ó°ÊÓ

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An ice-cube tray of negligible mass contains \(0.290 \mathrm{~kg}\) of water at \(18.0^{\circ} \mathrm{C}\). How much heat must be removed to cool the water to \(0.00^{\circ} \mathrm{C}\) and freeze it? Express your answer in joules, calories, and Btu.

Short Answer

Expert verified
First, calculate the heat removed to cool the water, then calculate the heat removed to freeze the water. Add up these heats to get the total heat removed which can then be converted into calories and Btu.

Step by step solution

01

Calculate Heat Removed to Cool the Water

To calculate the heat \(q\) removed to cool the water, use the specific heat equation: \(q = mc\Delta T\), where \(m = 0.290\, kg\) is the mass of the water, \(c = 4.186 \times 10^{3} \, J/(kg\cdot K)\) is the specific heat capacity of water, and \(\Delta T = 18.0 \, C - 0.00 \, C = 18.0 \, K\) is the change in temperature (K and C have the same intervals). So, \(q_1 = mc\Delta T = 0.290 \, kg \times 4.186 \times 10^{3} \, J/(kg\cdot K) \times 18.0 \, K\)
02

Calculate Heat Removed to Freeze the Water

To calculate the heat \(q\) removed to freeze the water, apply the heat of fusion equation \(q = mL\), where \(m = 0.290\, kg\) is the mass of the water and \(L = 3.33 \times 10^{5} \, J/kg\) is the latent heat of fusion. So, \(q_2 = mL = 0.290 \, kg \times 3.33 \times 10^{5} \, J/kg\)
03

Sum the Heats to Find the Total Heat Removed

Sum up \(q_1\) and \(q_2\) to get the total heat \(Q\) removed: \(Q = q_1 + q_2\)
04

Convert the Heat into Other Units

To convert joules to calories, use the conversion factor \(1 \, cal = 4.184 \, J\). To convert joules to Btu, use the conversion \(1 \, Btu = 1055 \, J\). Calculate the equivalent amounts of heat in calories and Btu.

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Key Concepts

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

Understanding the Specific Heat Equation
Learning how to calculate the amount of heat required to change the temperature of a substance is an essential principle in thermodynamics. The specific heat equation, represented as
\( q = mc\Delta T \),
is the mathematical expression that allows us to do this. In this equation, \( q \) is the amount of heat energy in joules, \( m \) is the mass of the substance in kilograms, \( c \) is the specific heat capacity (the amount of heat energy required to raise the temperature of 1 kilogram of the substance by 1 Kelvin), and \( \Delta T \) is the change in temperature in Kelvin. Kelvin and Celsius scales have the same intervals, so the change in Celsius can also be used.

It's important to note that the specific heat capacity is unique to each material, which means water will have a different value than say, iron or wood. This concept is crucial when solving problems involving heat transfer in various substances. In our given problem scenario, the equation helped calculate the exact amount of heat that needed to be removed to cool water down to its freezing point.
The Role of Latent Heat of Fusion in State Changes
When substances change state, for example, from liquid to solid, they either absorb or release energy. This energy is called latent heat. Specifically, the latent heat of fusion refers to the amount of energy required to change a substance from solid to liquid or vice versa without changing its temperature.
The equation used to calculate this is
\( q = mL \),
where \( q \) is the thermal energy change in joules, \( m \) is the mass of the substance in kilograms, and \( L \) is the latent heat of fusion specific to the substance, measured in joules per kilogram (J/kg). In the prior exercise, calculating the heat removed to freeze the water requires using water's specific latent heat of fusion, since the phase change happens at a constant temperature. Understanding this principle is integral when dealing with any process involving a phase transition, like refrigeration, melting or fabrication involving metals, or even in natural phenomena like snow formation.
Converting Energy Between Units
Different areas and professions measure energy in various units, so it's crucial to know how to convert between them. Common units for measuring heat energy include joules (J), calories (cal), and British Thermal Units (Btu). In our exercise, we converted the total heat removed from joules into calories and Btu using the conversion factors:
\(1 \text{ cal} = 4.184 \text{ J}\)
\(1 \text{ Btu} = 1055 \text{ J}\).
There are other energy units as well, such as kilowatt-hours (kWh) used in electricity measurements or therms used in natural gas measurement. Knowing these conversions is incredibly useful because it allows comparisons and calculations across different systems and standards. For example, when assessing energy consumption at home, you might need to convert your electricity usage to Btu or calories to understand it within a broader range of applications, like how much energy you're using compared to burning calories in physical exercise or the amount of heat your furnace produces in the winter.

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Most popular questions from this chapter

An electric kitchen range has a total wall area of \(1.40 \mathrm{~m}^{2}\) and is insulated with a layer of fiberglass \(4.00 \mathrm{~cm}\) thick. The inside surface of the fiberglass has a temperature of \(175^{\circ} \mathrm{C},\) and its outside surface is at \(35.0^{\circ} \mathrm{C}\). The fiberglass has a thermal conductivity of \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of \(1.40 \mathrm{~m}^{2}\) ? (b) What electric-power input to the heating element is required to maintain this temperature?

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass \(68 \mathrm{~kg}\) and surface area \(1.85 \mathrm{~m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{~W}, 80 \%\) of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\). (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right) ?\) (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (e) How many \(750 \mathrm{~mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of \(1.0 \mathrm{~kg}\).

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 \((95 \mathrm{~K}) ;\) (c) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{~K}\right)\)

A metal rod that is \(30.0 \mathrm{~cm}\) long expands by \(0.0650 \mathrm{~cm}\) when its temperature is raised from \(0.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\). A rod of a different metal and of the same length expands by \(0.0350 \mathrm{~cm}\) for the same rise in temperature. A third rod, also \(30.0 \mathrm{~cm}\) long, is made up of pieces of each of the above metals placed end to end and expands \(0.0580 \mathrm{~cm}\) between \(0.0^{\circ} \mathrm{C}\) and \(100.0^{\circ} \mathrm{C} .\) Find the length of each portion of the composite rod.

The molar heat capacity of a certain substance varies with temperature according to the empirical equation $$ C=29.5 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+\left(8.20 \times 10^{-3} \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}^{2}\right) T $$ How much heat is necessary to change the temperature of \(3.00 \mathrm{~mol}\) of this substance from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C} ?\) (Hint: Use Eq. (17.18) in the form \(d Q=n C d T\) and integrate. \()\)

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