/*! 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 12 The heat required to raise the t... [FREE SOLUTION] | 91Ó°ÊÓ

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The heat required to raise the temperature of \(m\) (kg) of a liquid from \(T_{1}\) to \(T_{2}\) at constant pressure is $$ Q=\Delta H=m \int_{T_{1}}^{T_{2}} C_{p}(T) d T $$ In high school and in first-year college physics courses, the formula is usually given as $$ Q=m C_{p} \Delta T=m C_{p}\left(T_{2}-T_{1}\right) $$ (a) What assumption about \(C_{p}\) is required to go from Equation 1 to Equation \(2 ?\) (b) The heat capacity \(\left(C_{p}\right)\) of liquid \(n\) -hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well- insulated vessel containing \(2.00 \mathrm{L}\) of liquid \(n-\mathrm{C}_{6} \mathrm{H}_{14}\) at \(T=300 \mathrm{K} .\) A combustion reaction known to release \(16.73 \mathrm{kJ}\) of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be \(3.10 \mathrm{K}\). In a separate experiment, it is found that \(6.14 \mathrm{kJ}\) of heat is required to raise the temperature of everything in the system except the hexane by \(3.10 \mathrm{K}\). Use these data to estimate \(C_{p}[\mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K})]\) for liquid \(n\) -hexane at \(T \approx 300 \mathrm{K},\) assuming that the condition required for the validity of Equation 2 is satisfied. Compare your result with a tabulated value.

Short Answer

Expert verified
The assumption required to go from Equation 1 to Equation 2 is that the heat capacity at constant pressure (\(C_p\)) is independent of temperature over the range \(T_1\) to \(T_2\). The specific heat capacity of liquid n-hexane at \(T \approx 300 K\) is estimated to be approximately \(298.2 J/(mol·K)\).

Step by step solution

01

Understand the relationship between the two equations

From the given equations, what transforms Equation 1 into Equation 2 is an assumption that the specific heat capacity at constant pressure (\(C_p\)) is constant within the temperature range considered (\(T_1\) to \(T_2\)). The quantity \(C_p\) is generally a function of temperature, but in many practical cases, it doesn’t vary significantly over the temperature change. Therefore, \(C_p\) can be taken out of the integral in Equation 1, simplifying into Equation 2.
02

Apply conservation of energy

Since there is no heat loss from the well-insulated calorimeter, all the heat \(Q\) generated by the combustion reaction goes into raising the temperature of the system contents. Let's denote the heat absorbed by n-hexane as \(Q_{n-C6H14}\), and the heat absorbed by the rest of the system as \(Q_{rest} = 6.14 kJ\). According to conservation of energy, \(Q = Q_{n-C6H14} + Q_{rest}\). The total heat generated by the combustion \(Q\) is given as \(16.73 kJ\). Therefore, we can derive \(Q_{n-C6H14} = Q - Q_{rest} = 16.73 kJ - 6.14 kJ = 10.59 kJ\).
03

Calculate C_p of n-hexane

For the n-hexane liquid, \(C_p\left(n-C6H14\right)\) can be calculated using Equation 2: \(Q_{n-C6H14} = mC_p\Delta T\). We know the mass of hexane \(m = 2.00 L * 0.659 kg/L = 1.318 kg\), where 0.659 kg/L is the density of n-hexane. And the temp. difference \(\Delta T = 3.1 K\). Substituting these values into Equation 2, we can solve for \(C_p = Q_{n-C6H14} / (m\Delta T) = 10.59 kJ / (1.318 kg * 3.1 K) = 2.57 kJ/kg·K\).
04

Convert into standard units

In chemistry, the standard unit for heat capacity is J/(mol·K). Considering 1KJ = 1000J, and the molar mass of n-hexane is 86.18 g/mol, we can convert the specific heat capacity into standard units \[C_p = 2.57 KJ/kg·K * 1000 J/KJ / (86.18 g/mol / 1000 g/kg) = 298.2 J/(mol·K)\].

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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, denoted as \(C_p\) when at constant pressure, is a critical concept in understanding how substances heat up or cool down. It is defined as the amount of heat required to raise the temperature of one unit of mass of a substance by one degree Celsius (or one Kelvin). This property is unique to each substance and depends on its molecular structure and interactions.When we talk about specific heat capacity in practical scenarios, it is often assumed to be constant over a given temperature range. However, \(C_p\) can vary slightly depending on the temperature. For problems ensuring or needing precision, it's crucial to acknowledge these variations. In basic thermodynamics, the simplified equation \(Q = m C_p \Delta T\) is used, assuming \(C_p\) does not change over the temperature range \(T_1\) to \(T_2\). This equation helps in calculating the heat (Q) required or released, reinforcing the importance of specific heat capacity in thermodynamic equations.
  • Specific heat capacity is often kept constant for simplicity in calculations.
  • Varies slightly with temperature but often negligible for small ranges.
  • Essential in calculating the heat transfer during a temperature change.
Calorimetry
Calorimetry is the science of measuring the heat of chemical reactions or physical changes. A calorimeter is the device used for this purpose, and it measures the heat exchanged with the surroundings. In a bomb calorimeter experiment like the exercise described, the key idea is that the calorimeter is well-insulated so that no heat is lost or gained from the surroundings.In our problem, a combustion reaction occurs in the bomb, releasing a known amount of heat. This heat raises the temperature of both the hexane and the rest of the calorimeter system. By knowing the heat released by the reaction and measuring the change in temperature, we can calculate the specific heat capacity of the substance reacting—in this case, n-hexane.
  • Ensures no heat is exchanged with surroundings for accurate measurements.
  • Combines known reaction heat with temperature change to find \(C_p\).
  • Used extensively in determining thermal properties of substances.
Combustion Reaction
A combustion reaction is a high-energy chemical reaction between a fuel and an oxidant, producing heat and often light. In the context of our calorimetry exercise, combustion serves as the source of heat. A controlled combustion reaction takes place inside the calorimeter, releasing a predetermined amount of thermal energy. The heat produced by this reaction causes a temperature rise in the calorimeter's contents. For the specific reaction discussed in this problem, 16.73 kJ of heat is released, which then evenly distributes between the hexane and the residual calorimeter components. By partitioning the heat absorbed and considering the temperature change, we can accurately measure the specific heat capacity of the n-hexane present.
  • Involves rapid oxidation of a fuel, releasing energy.
  • Heat from combustion used in estimating specific heat capacities.
  • Central to many industrial processes and calorimetric studies.

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Most popular questions from this chapter

The heat capacities of a substance have been defined as $$C_{v}=\left(\frac{\partial \hat{U}}{\partial T}\right)_{V}, \quad C_{p}=\left(\frac{\partial \hat{H}}{\partial T}\right)_{P}$$ Use the defining relationship between \(\hat{H}\) and \(\hat{U}\) and the fact that \(\hat{H}\) and \(\hat{U}\) for ideal gases are functions only of temperature to prove that \(C_{p}=C_{v}+R\) for an ideal gas (Eq. \(8.3-12\) ).

Fish and wildlife managers have determined that a sudden temperature increase greater than \(5^{\circ} \mathrm{C}\) would be harmful to the marine ecosystem of a river. Warmer waters contain less dissolved oxygen and cause organisms in a river to increase their metabolism; if the temperature increase is sudden, the organisms do not have time to adapt to the new environment and likely will die. (Changes in river temperatures of five degrees and more due to seasonal temperature variations are common, but those temperature changes are gradual.) A proposed chemical plant plans to use river water for process cooling. The river flows at a rate of \(15.0 \mathrm{m}^{3} / \mathrm{s}\) at a temperature of \(15^{\circ} \mathrm{C}\), and a fraction of it will be diverted to the plant. Preliminary calculations reveal that the cooling water will remove \(5.00 \times 10^{5} \mathrm{kJ} / \mathrm{s}\) of heat from the plant. A portion of the extracted water will evaporate from the plant into the atmosphere, and the remainder will be returned to the river at a temperature of \(35^{\circ} \mathrm{C}\). (a) Draw and completely label a flowchart of the process and prove that there is enough information available to calculate all of the unknown stream flow rates on the chart. (b) Estimate the fraction of the river flow that must be diverted to the plant and the percentage of the cooling water that evaporates. Assume that water has a constant heat capacity of \(4.19 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) and a heat of vaporization roughly that of water at the normal boiling point, and also assume that the specific enthalpy of the water vapor relative to liquid water at \(15^{\circ} \mathrm{C}\) equals the heat of vaporization. (c) Write (but don't evaluate) an expression for the enthalpy change neglected by the assumption about the specific enthalpy of the steam.

The off-gas from a reactor in a process plant in the heart of Freedonia has been condensing and plugging up the vent line, causing a dangerous pressure buildup in the reactor. Plans have been made to send the gas directly from the reactor into a cooling condenser in which the gas and liquid condensate will be brought to \(25^{\circ} \mathrm{C}.\) (a) You have been called in as a consultant to aid in the design of this unit. Unfortunately, the chief (and only) plant engineer has disappeared and nobody else in the plant can tell you what the offgas is (or what anything else is, for that matter). However, a job is a job, and you set out to do what you can. You find an elemental analysis in the engineer's notebook indicating that the gas formula is \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\). On another page of the notebook, the off-gas flow rate is given as \(235 \mathrm{m}^{3} / \mathrm{h}\) at \(116^{\circ} \mathrm{C}\) and 1 atm. You take a sample of the gas and cool it to \(25^{\circ} \mathrm{C}\), where it proves to be a solid. You then heat the solidified sample at 1 atm and note that it melts at \(52^{\circ} \mathrm{C}\) and boils at \(113^{\circ} \mathrm{C}\). Finally, you make several assumptions and estimate the heat removal rate in \(\mathrm{kW}\) required to bring the off-gas from \(116^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\). What is your result? (b) If you had the right equipment, what might you have done to get a better estimate of the cooling rate?

Among the best-known building blocks in nanotechnology applications are nanoparticles of noble metals. For example, colloidal suspensions of silver or gold nanoparticles (10-200 nm) exhibit vivid colors because of intense optical absorption in the visible spectrum, making them useful in colorimetric sensors. In the illustration shown below, a suspension of gold nanoparticles of a fairly uniform size in water exhibits peak absorption near a wavelength of \(525 \mathrm{nm}\) (near the blue region of the visible spectrum of light). When one views the solution in ambient (white) light, the solution appears wine-red because the blue part of the spectrum is largely absorbed. When the nanoparticles aggregate to form large particles, an optical absorption peak near \(600-700 \mathrm{nm}\) (near the red region of the visible spectrum) is observed. The breadth of the peak reflects a fairly broad particle size distribution. The solution appears bluish because the unabsorbed light reaching the eye is dominated by the short (blue-violet) wavelength region of the spectrum. since the optical properties of metallic nanoparticles are a strong function of their size, achieving a narrow particle size distribution is an important step in the development of nanoparticle applications. A promising way to do so is laser photolysis, in which a suspension of particles of several different sizes is irradiated with a high-intensity laser pulse. By carefully selecting the wavelength and energy of the pulse to match an absorption peak of one of the particle sizes (e.g., irradiating the red solution in the diagram with a \(525 \mathrm{nm}\) laser pulse), particles of or near that size can be selectively vaporized. (a) A spherical silver nanoparticle of diameter \(D\) at \(25^{\circ} \mathrm{C}\) is to be heated to its normal boiling point and vaporized with a pulsed laser. Considering the particle a closed system at constant pressure, write the energy balance for this process, look up the physical properties of silver that are required in the energy balance, and perform all the required substitutions and integrations to derive an expression for the energy \(Q_{\text {abs }}(\mathrm{J})\) that must be absorbed by the particle as a function of \(D(\mathrm{nm})\) (b) The total energy absorbed by a single particle \(\left(Q_{\text {abs }}\right)\) can also be calculated from the following relation: $$ Q_{\mathrm{abs}}=F A_{\mathrm{p}} \sigma_{\mathrm{abs}} $$ where \(F\left(\mathrm{J} / \mathrm{m}^{2}\right)\) is the energy in a single laser pulse per unit spot area (area of the laser beam) and \(A_{\mathrm{p}}\left(\mathrm{m}^{2}\right)\) is the total surface area of the nanoparticle. The effectiveness factor, \(\sigma_{\mathrm{ahs}},\) accounts for the efficiency of absorption by the nanoparticle at the wavelength of the laser pulse and is dependent on the particle size, shape, and material. For a spherical silver nanoparticle irradiated by a laser pulse with a peak wavelength of \(532 \mathrm{nm}\) and spot diameter of \(7 \mathrm{mm}\) with \(D\) ranging from 40 to \(200 \mathrm{nm}\), the following empirical equation can be used for \(\sigma_{\mathrm{abs}}\) $$ \sigma_{\mathrm{abs}}=\frac{1}{4}\left[0.05045+2.2876 \exp \left(-\left(\frac{D-137.6}{41.675}\right)^{2}\right)\right] $$ where \(\sigma_{\text {abs }}\) and the leading \(\frac{1}{4}\) are dimensionless and \(D\) has units of nm. Use the results of Part (a) to determine the minimum values of F required for complete vaporization of single nanoparticles with diameters of \(40.0 \mathrm{nm}, 80.0 \mathrm{nm},\) and \(120.0 \mathrm{nm}\). If the pulse frequency of the laser is \(10 \mathrm{Hz}\) (i.e., 10 pulses per second), what is the minimum laser power \(P(\mathrm{W})\) required for each of those values of D? (Hint: Set up a dimensional equation relating \(P\) to \(F\).) (c) Suppose you have a suspension of a mixture of \(D=40 \mathrm{nm}\) and \(D=120 \mathrm{nm}\) spherical silver nanoparticles and a \(10 \mathrm{Hz} / 532 \mathrm{nm}\) pulsed laser source with a \(7 \mathrm{nm}\) diameter spot and adjustable power. Describe how you would use the laser to produce a suspension of particles of only a single size and state what that size would be.

The specific internal energy of formaldehyde (HCHO) vapor at 1 atm and moderate temperatures is given by the formula $$\hat{U}(\mathrm{J} / \mathrm{mol})=25.96 T+0.02134 T^{2}$$ where \(T\) is in \(^{\circ} \mathrm{C}\) (a) Calculate the specific internal energies of formaldehyde vapor at \(0^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\). What reference temperature was used to generate the given expression for \(\hat{U} ?\) (b) The value of \(\hat{U}\) calculated for \(200^{\circ} \mathrm{C}\) is not the true value of the specific internal energy of formaldehyde vapor at this condition. Why not? (Hint: Refer back to Section 7.5a.) Briefly state the physical significance of the calculated quantity. (c) Use the closed system energy balance to calculate the heat (J) required to raise the temperature of 3.0 mol HCHO at constant volume from 0^0 C to 200^'C. List all of your assumptions. (d) From the definition of heat capacity at constant volume, derive a formula for \(C_{v}(T)\left[\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\) Then use this formula and Equation \(8.3-6\) to calculate the heat \((\) J) required to raise the temperature of 3.0 mol of HCHO(v) at constant volume from 0^ C to 200^'C. [You should get the same result you got in Part (c).]

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