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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·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°C or 300 K.
  • The final temperature, \(T_f\), is the end point of the process, which in this exercise is 227°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

One end of an insulated metal rod is maintained at 100.0\(^\circ\)C, and the other end is maintained at 0.00\(^\circ\)C by an ice-water mixture. The rod is 60.0 cm long and has a cross-sectional area of 1.25 cm\(^2\). The heat conducted by the rod melts 8.50 g of ice in 10.0 min. Find the thermal conductivity \(k\) of the metal.

A glass flask whose volume is 1000.00 cm\(^3\) at 0.0\(^\circ\)C is completely filled with mercury at this temperature. When flask and mercury are warmed to 55.0\(^\circ\)C, 8.95 cm\(^3\) of mercury overflow. If the coefficient of volume expansion of mercury is \(18.0 \times 10{^-}{^5} K{^-}{^1}\), compute the coefficient of volume expansion of the glass.

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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 \(2.42 \times 10{^6} J/kg\).)

A carpenter builds a solid wood door with dimensions 2.00 m \(\times\) 0.95 m \(\times\) 5.0 cm. Its thermal conductivity is k = 0.120 W/m \(\cdot\) K. The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.8-cm thickness of solid wood. The inside air temperature is 20.0\(^\circ\)C, and the outside air temperature is -8.0\(^\circ\)C. (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 m on a side is inserted in the door? The glass is 0.450 cm thick, and the glass has a thermal conductivity of 0.80 W/m \(\cdot\) K. The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cm of glass.

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