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Chapter 17: Problem 88

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

Short Answer

Expert verified
The heat necessary is 22107 J.

Step by step solution

01

Identify the Variables

We begin by identifying the given variables:- Initial temperature, \( T_i = 27^\circ C = 300 \, K \)- Final temperature, \( T_f = 227^\circ C = 500 \, K \)- Number of moles, \( n = 3.00 \, \text{mol} \)- Molar heat capacity, \( C = 29.5 \, \text{J/mol} \cdot \text{K} + (8.20 \times 10^{-3} \, \text{J/mol} \cdot \text{K}^2)T \)
02

Setup the Heat Transfer Equation

Using the hint provided, we know that the differential form of the heat transfer can be expressed as \( dQ = nC dT \). This tells us how much heat is added when the temperature changes by an infinitesimal amount \( dT \).
03

Substitute the Molar Heat Capacity into the Equation

Replace \( C \) with the given function: \[dQ = 3.00 \left(29.5 + (8.20 \times 10^{-3})T\right) dT\]This modifies \( dQ \) in terms of \( T \).
04

Integrate the Equation

Calculate the total heat \( Q \) by integrating \( dQ \) from \( T_i = 300 \, K \) to \( T_f = 500 \, K \):\[Q = \int_{300}^{500} 3.00 \left(29.5 + (8.20 \times 10^{-3})T\right) dT\]Solving the integral:\[Q = 3.00 \left[29.5T + \frac{8.20 \times 10^{-3}}{2} T^2 \right]_{300}^{500}\]
05

Evaluate the Integral

Evaluate the integrated function at the bounds:\[Q = 3.00 \left[\left(29.5 \times 500 + \frac{8.20 \times 10^{-3}}{2} \times 500^2\right) - \left(29.5 \times 300 + \frac{8.20 \times 10^{-3}}{2} \times 300^2\right)\right]\]This simplifies to:\[Q = 3.00 \left[14750 + 1.025 \times 500^2 - (8850 + 1.025 \times 300^2)\right]\]
06

Calculate the Numerical Value

Calculate each term inside the brackets:- \( 29.5 \times 500 = 14750 \)- \( (8.20 \times 10^{-3}/2) \times 500^2 = 1025 \)- \( 29.5 \times 300 = 8850 \)- \( (8.20 \times 10^{-3}/2) \times 300^2 = 369 \)Substitute these values back into the equation to find \( Q \):\[Q = 3.00 \left((14750 + 1025) - (8850 + 369)\right)\]\[Q = 3.00 \times 7369 = 22107 \, \text{J}\]
07

Conclusion

The total heat required to change the temperature of the substance is 22107 J.

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Key Concepts

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

Molar Heat Capacity
Molar heat capacity is an important concept in thermodynamics as it measures the amount of heat a substance can hold per mole for a given temperature change. Expressing it as a function of temperature, like in the provided equation, is common for many substances. This equation describes how the heat capacity changes depending on the temperature, combining a constant term and a temperature-dependent term:
  • The constant term (29.5 J/mol路K) represents the basic heat absorption ability inherent to the substance, regardless of temperature.
  • The temperature-dependent term \(8.20 \times 10^{-3}T\) reflects how the capacity increases or decreases as the substance's temperature changes.

In this way, molar heat capacity is crucial for determining how much heat is needed during a temperature change, thus playing a central role when calculating heat transfer in varying thermal conditions.
Temperature Change
Temperature change is simply the difference between the initial and final temperatures of a substance when it undergoes heating or cooling. In thermodynamic calculations, temperature is typically converted from Celsius to Kelvin for accuracy.
  • The initial temperature, \(T_i\), is where the process starts; for this exercise, it is 27掳C or 300 K.
  • The final temperature, \(T_f\), is the end point of the process, which in this exercise is 227掳C or 500 K.

The temperature change, \(\Delta T\), is the difference \(T_f - T_i = 500 \,K - 300 \,K = 200 \,K\). This change is a driving force in the thermodynamic equations, affecting the amount of heat transfer required to achieve that change.
Heat Transfer
Heat transfer is the movement of thermal energy from one place to another. In a thermal system, it is often calculated using the relation \(dQ = nC dT\), where \(dQ\) is the differential heat added, \(n\) is the number of moles, \(C\) is the molar heat capacity, and \(dT\) is the small change in temperature. This relation tells us how heat added depends on the thermal properties of the substance and the current temperature.
  • In our case, the number of moles, \(n = 3.00\), affects how much total heat is required.
  • The molar heat capacity, expressed as \(29.5 + (8.20 \times 10^{-3})T\), is substituted into this formula, creating a differential equation ready to be integrated.

This calculated heat transfer quantifies the energy needed to achieve the desired temperature change, playing a vital role in thermal management and engineering.
Integration in Thermodynamics
Integration in thermodynamics allows for calculating the total heat transfer over a finite range of temperatures. Starting with the expression \(dQ = nC dT\), we integrate both sides from the initial temperature \(T_i\) to the final temperature \(T_f\).
  • This involves integrating the molar heat capacity equation over the temperature range. Here, \(Q = \int_{300}^{500} 3.00(29.5 + (8.20 \times 10^{-3})T) dT\).
  • The integral evaluates to \(Q = 3.00 \left[29.5T + \frac{8.20 \times 10^{-3}}{2} T^2 \right]_{300}^{500}\).

Solving this integral provides the total heat required, using fundamental calculus that bridges changes in state functions across temperature ranges. This method is essential for accurately determining energy exchanges in chemistry and physics.

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Most popular questions from this chapter

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0\(^\circ\)C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0\(^\circ\)C? You can ignore the heat transferred to the container.

Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to 34.0\(^\circ\)C overnight and rise to 40.0\(^\circ\)C during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400-kg camel would have to drink if it attempted to keep its body temperature at a constant 34.0\(^\circ\)C by evaporation of sweat during the day (12 hours) instead of letting it rise to 40.0\(^\circ\)C. (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 J/kg \(\cdot\) K. The heat of vaporization of water at 34\(^\circ\)C is \(2.42 \times 10{^6} J/kg\).)

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a 70.0-kg man to cool his body 1.00 C\(^\circ\)? The heat of vaporization of water at body temperature (37\(^\circ\)C) is \(2.42 \times 10{^6} J/kg\). The specific heat of a typical human body is 3480 J/kg \(\cdot\) K (see Exercise 17.25). (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can (355 cm\(^3\)).

The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the surface area of the filament of a 150-W bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves. (Only a fraction of the radiation appears as visible light.)

(a) A typical student listening attentively to a physics lecture has a heat output of 100 W. How much heat energy does a class of 140 physics students release into a lecture hall over the course of a 50-min lecture? (b) Assume that all the heat energy in part (a) is transferred to the 3200 m\(^3\) of air in the room. The air has specific heat 1020 J/kg \(\cdot\) K and density 1.20 kg/m\(^3\). If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50-min lecture? (c) If the class is taking an exam, the heat output per student rises to 280 W. What is the temperature rise during 50 min in this case?
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