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Chapter 17: Problem 33

An aluminum tea kettle with mass 1.50 \(\mathrm{kg}\) and containing 1.80 \(\mathrm{kg}\) of water is placed on a stove. If no heat is lost to the surroundings, how much heat must be added to raise the temperature from \(20.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The total heat required is 615.6 kJ.

Step by step solution

01

Identify Given Information

We have an aluminum kettle with a mass of 1.50 kg and water inside it with a mass of 1.80 kg. We need to raise the temperature from 20.0掳C to 85.0掳C.
02

Determine Specific Heat Capacities

The specific heat capacity of aluminum is 0.900 J/g掳C, and for water, it is 4.186 J/g掳C.
03

Convert Mass to Grams

Convert the mass from kilograms to grams for consistency with the specific heat capacity units: 1.50 kg of aluminum = 1500 g 1.80 kg of water = 1800 g.
04

Calculate Heat Required for Aluminum

Use the formula for heat: \[ q = m \, \times \, c \, \times \, \Delta T \]where: - \( q \) is the heat added,- \( m \) is the mass,- \( c \) is the specific heat capacity,- \( \Delta T \) is the change in temperature.For aluminum:\[ q_{\text{aluminum}} = 1500 \, \text{g} \, \times \, 0.900 \, \text{J/g掳C} \, \times \, (85.0掳C - 20.0掳C) \]\[ q_{\text{aluminum}} = 1500 \, \text{g} \, \times \, 0.900 \, \text{J/g掳C} \, \times \, 65.0掳C \]Calculate the result.
05

Calculate Heat Required for Water

Apply the same formula for water:\[ q_{\text{water}} = 1800 \, \text{g} \, \times \, 4.186 \, \text{J/g掳C} \, \times \, (85.0掳C - 20.0掳C) \]\[ q_{\text{water}} = 1800 \, \text{g} \, \times \, 4.186 \, \text{J/g掳C} \, \times \, 65.0掳C \]Calculate the result.
06

Calculate Total Heat Required

Add the heat required for the aluminum and the water to find the total heat required:\[ q_{\text{total}} = q_{\text{aluminum}} + q_{\text{water}} \] Substitute in the values from the previous steps and calculate the total heat.

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fascinating property of materials. It tells us how much heat energy is needed to raise the temperature of one gram of a substance by one degree Celsius. In this exercise, we deal with two different substances: aluminum and water.
Aluminum has a specific heat capacity of 0.900 J/g掳C, while water holds a much higher value of 4.186 J/g掳C. This means that water requires more heat energy than aluminum for the same mass and temperature change.
Understanding specific heat capacity helps us determine how different substances will react to heat. It allows us to calculate precisely how much energy is needed to reach a desired temperature change.
Thermal Energy Calculation
In thermal energy calculations, we use a straightforward formula to find the amount of heat energy required:
  • \( q = m \times c \times \Delta T \)
Here, \( q \) represents the heat added, \( m \) is the mass in grams, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.
In the problem at hand, we calculate the heat required for both the aluminum kettle and the water inside. For example, to heat the aluminum, we use its mass (1500 g), its specific heat capacity (0.900 J/g掳C), and the temperature change (65掳C).
Applying these values gives us an estimate of the energy needed, which we can similarly do for the water. This straightforward process enables clear assessments of the energy demands involved in heating different substances.
Temperature Change
Temperature change is at the heart of any thermal energy problem. It represents the difference between the initial and final temperatures of the substances in question.
In this scenario, the temperature change \( \Delta T \) is from 20.0掳C to 85.0掳C. To find \( \Delta T \), we simply subtract the initial temperature from the final temperature:
  • \( \Delta T = 85.0掳C - 20.0掳C = 65.0掳C \)
This result tells us how much each substance's temperature needs to rise. Whether working with metals like aluminum, or liquids such as water, knowing the temperature change is crucial for calculating the total thermal energy required. These calculations rely on the specific properties of each material, emphasizing how important it is to understand those variations across different substances.

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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 \(\mathrm{m}^{2} ?(\mathrm{b})\) What electric- power input to the heating element is required to maintain this temperature?

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?

Food Intake of a Hamster. The energy output of an animal engaged in an activity is called the basal metabolic rate (BMR) and is a measure of the conversion of food energy into other forms of energy. A simple calorimeter to measure the BMR consists of an insulated box with a thermometer to measure the temperature of the air. The air has density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and specific heat 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . A 50.0 -g hamster is placed in a calorimeter that contains 0.0500 \(\mathrm{m}^{3}\) of air at room temperature. (a) When the hamster is running in a wheel, the temperature of the air in the calorimeter rises 1.60 \(\mathrm{C}^{\circ}\) per hour. How much heat does the running hamster generate in an hour? Assume that all this heat goes into the air in the calorimeter. You can ignore the heat that goes into the walls of the box and into the thermometer, and assume that no heat is lost to the surroundings. (b) Assuming that the hamster converts seed into heat with an efficiency of 10\(\%\) and that hamster seed has a food energy value of 24 \(\mathrm{J} / \mathrm{g}\) , how many grams of seed must the hamster eat per hour to supply this energy?

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.

A U.S. penny has a diameter of 1.9000 \(\mathrm{cm}\) at \(20.0^{\circ} \mathrm{C} .\) The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10^{-5} \mathrm{K}^{-1} .\) What would its diameter be on a hot day in Death Valley \(\left(48.0^{\circ} \mathrm{C}\right) ?\) On a cold night in the mountains of Grecnland \(\left(-53^{\circ} \mathrm{C}\right) ?\)
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