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Chapter 18: Problem 32

The specific heat of a substance varies with temperature according to the function \(c=0.20+0.14 T+0.023 T^{2}\), with \(T\) in \({ }^{\circ} \mathrm{C}\) and \(c\) in cal \(/ \mathrm{g} \cdot \mathrm{K}\). Find the energy required to raise the temperature of \(2.0 \mathrm{~g}\) of this substance from \(5.0^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The energy required is approximately 81.96 calories.

Step by step solution

01

Understanding the Formula

The specific heat of the substance is given by the equation \( c = 0.20 + 0.14T + 0.023T^2 \). This means that the specific heat changes with temperature, \( T \), measured in Celsius. Our goal is to find out how much energy it takes to increase 2 grams of the substance's temperature from 5掳C to 15掳C using this equation.
02

Integrating the Specific Heat

We need to find the energy used, which can be calculated by integrating the specific heat from the initial temperature \( T_i = 5 \)掳C to the final temperature \( T_f = 15 \)掳C. The energy \( Q \) required is given by the formula:\[Q = m \times \int_{T_i}^{T_f} c(T) \, dT\]where \( m = 2.0 \) g is the mass of the substance.
03

Setting Up the Integral

Insert the expression for \( c(T) \) into the integral:\[Q = 2.0 \times \int_{5}^{15} \left(0.20 + 0.14T + 0.023T^2\right) \, dT\]
04

Calculating the Integral Components

Calculate the integral:- \( \int 0.20 \, dT = 0.20T \)- \( \int 0.14T \, dT = 0.07T^2 \)- \( \int 0.023T^2 \, dT = 0.00767T^3 \)So:\[Q = 2.0 \times \left[0.20T + 0.07T^2 + 0.00767T^3\right]_5^{15}\]
05

Evaluating the Integral Limits

Evaluate the expression at the limits 5 and 15:- At \( T = 15 \): \( 0.20(15) + 0.07(15)^2 + 0.00767(15)^3 = 3 + 15.75 + 25.9375 = 44.6875 \)- At \( T = 5 \): \( 0.20(5) + 0.07(5)^2 + 0.00767(5)^3 = 1 + 1.75 + 0.9575 = 3.7075 \)Subtract the results: \( 44.6875 - 3.7075 = 40.98 \)
06

Calculating the Final Energy

Multiply by the mass of the substance:\[Q = 2.0 \times 40.98 = 81.96 \text{ cal}\]
07

Conclusion

The energy required to increase the temperature of 2.0 grams of the substance from 5掳C to 15掳C is approximately 81.96 calories.

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Key Concepts

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

Thermodynamics
The branch of physics known as thermodynamics explores how heat, energy, and work interact. It examines the energy transformations that occur in any system, especially during chemical reactions or changes in physical state.

Understanding specific heat and its variation with temperature is crucial in thermodynamics. This concept describes the relationship between heat energy added to a substance and the resultant temperature change. A high specific heat indicates that more energy is needed to change the temperature of a substance, while a low specific heat means less energy is required.
  • **Heat energy transfer**: In this exercise, the goal is to understand the energy transfer necessary to raise the temperature of a substance.
  • **Variable specific heat**: We've been given a function, illustrating that the specific heat changes with temperature, indicating a more complex relationship than constant specific heat.

As we explore thermodynamics, it's essential to consider these variations because they impact the energy calculations and provide insight into how microscopic interactions translate into macroscopic phenomena.
Calorimetry
Calorimetry is a scientific method used to measure the heat transferred to or from a substance. It's an essential aspect of thermodynamics and focuses on understanding heat exchanges during physical changes or chemical reactions.

In this exercise, calorimetry aids us by providing a framework to calculate the energy required to change a substance's temperature.
  • **Heat capacity**: In a constant specific heat scenario, a simple multiplication would suffice; however, here it necessitates integration due to its dependency on temperature.
  • **Practical applications**: Calorimetry is widely used in various fields, including chemistry, biology, and physics, to measure reaction enthalpies and specific heats.

The knowledge of how much energy is absorbed or released by a substance is fundamental in calorimetry. It offers a quantitative measure of the energetic changes occurring within the system. This exercise, with its variable specific heat, prepares one for real-world applications where such variations are common.
Integral Calculus
Integral calculus allows us to find areas, volumes, central points, and many other useful things. In the context of our exercise, it provides the necessary tools to solve the problem of variable specific heat by integrating the function over a specific interval.

  • **Integration process**: This involves setting up the integral of the specific heat function with respect to temperature. Since specific heat varies, a definite integral over the temperature range is evaluated.
  • **Integral evaluation**: Integrating the given function from the initial to the final temperature gives us the total energy required to achieve the temperature change.

By integrating, we are essentially summing up small differences over the interval, which provides the total energy absorbed by the substance. Integral calculus thus serves as a powerful tool in thermodynamics and calorimetry for handling complex temperature-dependent scenarios.

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Most popular questions from this chapter

To withstand the harsh weather of the Antarctic, emperor penguins huddle in groups (Fig. \(18-50)\). Assume that a penguin is a circular cylinder with a top surface area \(a=0.34 \mathrm{~m}^{2}\) and height \(h=1.1 \mathrm{~m}\). Let \(P_{r}\) be the rate at which an individual penguin radiates energy to the environment (through the top and the sides); thus \(N P_{r}\) is the rate at which \(N\) identical, wellseparated penguins radiate. If the penguins huddle closely to form a huddled cylinder with top surface area \(N a\) and height \(h\), the cylinder radiates at the rate \(P_{h}\). If \(N=1000\), (a) what is the value of the fraction \(P_{h} / N P_{r}\) and (b) by what percentage does huddling reduce the total radiation loss?

What mass of butter, which has a usable energy content of \(6.0 \mathrm{Cal} / \mathrm{g}(=6000 \mathrm{cal} / \mathrm{g})\), would be equivalent to the change in gravitational potential energy of a \(73.0 \mathrm{~kg}\) man who ascends from sea level to the top of Mt. Everest, at elevation \(8.84 \mathrm{~km}\) ? Assume that the average \(g\) for the ascent is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

Suppose the temperature of a gas is \(373.15 \mathrm{~K}\) when it is at the boiling point of water. What then is the limiting value of the ratio of the pressure of the gas at that boiling point to its pressure at the triple point of water? (Assume the volume of the gas is the same at both temperatures.)

A cold beverage can be kept cold even on a warm day if it is slipped into a porous ceramic container that has been soaked in water. Assume that energy lost to evaporation matches the net energy gained via the radiation exchange through the top and side surfaces. The container and beverage have temperature \(T=15^{\circ} \mathrm{C}\), the environment has temperature \(T_{\mathrm{env}}=32^{\circ} \mathrm{C}\), and the container is a cylinder with radius \(r=2.2 \mathrm{~cm}\) and height \(10 \mathrm{~cm}\). Approximate the emissivity as \(\varepsilon=1\), and neglect other energy exchanges. At what rate \(d m / d t\) is the container losing water mass?

A \(150 \mathrm{~g}\) copper bowl contains \(220 \mathrm{~g}\) of water, both at \(20.0^{\circ} \mathrm{C}\). A very hot \(300 \mathrm{~g}\) copper cylinder is dropped into the water, causing the water to boil, with \(5.00 \mathrm{~g}\) being converted to steam. The final temperature of the system is \(100^{\circ} \mathrm{C}\). Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder?
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