Thermodynamics and Temperature Measurements - Assignment Example

Thermodynamics 1) Calculate ΔS for the following reaction, using the thermodynamic data provided. a. NaCl(s) → Na+(aq) + Cl¯(aq) The enthalpy of hydration ΔH = +3.9 JK−1mol−1 ΔS­­­sys ­­ = +39.2 JK−1mol−1 ΔS­­­surr­­ =- ΔH/T = - (+39.2 JK−1mol−1 )/ 298K = -13.1 JK−1mol−1 ΔS­­­total ­­= ΔS­­­sys + ΔS­­­surr­­ = 39.2 – 13.1 = 26.1 JK−1mol−1 b. 2NO(g)+O2(g) → N2O4(g), The standard molar entropies of reactants and products are So (NO(g)) = 211.2 JK−1mol−1 So (O2(g)) = 205.1 JK−1mol−1 So (N2O4((g)) = 304.3 JK−1mol−1 ∆ So = Σ So(Products) - Σ So(Reactants) ∆ So = So (N2O4((g)) –2So (NO(g)) – So (O2(g)) =304.3 JK−1mol−1 –422.4 JK−1mol−1 – 205.1 JK−1mol−1 =-323.2 JK−1mol−1 c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). The standard molar entropies of reactants and products are So (CH4) = 197.6 JK−1mol−1 So (O2) = 205.138 Jmol−1K−1 So (CO2) = 213.74 Jmol−1K−1 So (H2O (l)) = 69.91 J/K mol ∆ So = Σ So(Products) - Σ So(Reactants) ∆ So = So (CO2) +2 So (H2O (l)) – So (CH4) – 2So (O2) ∆ So =213.74JK−1mol−1 +139.82JK−1mol−1 - 197.6 JK−1mol−1 – 410.276 JK−1mol−1 Therefore ∆ So = -254.316JK−1mol−1 d. 2NO2(g) ↔ N2O4(g) The standard molar entropies of reactants and products are So (NO2(g)) = 240.1 JK−1mol−1 So (N2O4((g)) = 304.3 JK−1mol−1 ∆ So = Σ So(Products) - Σ So(Reactants) ∆ So = So (N2O4((g)) –2So (NO2(g)) =304.3 JK−1mol−1 –480.2 JK−1mol−1 =-175.9 JK−1mol−1 2. These questions test your understanding of temperature measurements and temperature scales. a. Body temperature is 37°C what is this in Kelvin, Fahrenheit and Rankine scales? To convert from Celsius scale to Kelvin scale we add a value of 273.15, so the temperature of the human body in the Kelvin scale is 37 + 273.15 = 310.15K To convert from Celsius to Fahrenheit scale we multiply the value by 9/5 and add 32˚ Therefore, 37°C = {[37x(9/5)] + 32} ˚F = 98.6 ˚F To obtain temperature in Rankine scale we add 459.67˚ to the temperature measure in Fahrenheit scale. So we have 98.60 F = 98.6+ 459.67 = 558.270 R Hence a temperature of 37°C is equivalent to 310.15 K, 98.60F or 558.270R b. What is absolute zero in Celsius, Fahrenheit and Rankine scales? Absolute zero on the Kelvin scale is 0K, so on the Celsius scale it is (0 -273.15) °C, which is -273.15°C On the Fahrenheit scale this is equal to {[-273.15x(9/5)] + 32} ˚F = -459.67 ˚F On the Rankine scale absolute temperature is defined at 0R. Rankine is to Fahrenheit as Kelvin is to Celsius c. The temperature of a system rises by 45°C during a heating process. Express this rise in temperature in Kelvins. The difference of temperatures on Kelvin scale is equal to the difference of converted equivalent temperatures on the Celsius scale. Hence a 45°C rise is equivalent to a 45K rise in the temperature. d. The temperature of a system rises by 180°F during a heating process. Express this rise in temperature in R, K and °C. The Fahrenheit and the Rankine scales are equally spaced with respect to the degrees. So a 180°F rise is equivalent to a 180 R rise in temperature To convert difference of temperatures on Fahrenheit scale to Celsius scale we multiply by 5/9. So 180F° = (180×5/9) = 100C˚ Similarly the Celsius and the Kelvin scales are equally spaced with respect to the degrees. So a 100˚C rise in temperature is equivalent to a rise of 100 kelvin rise. 3) What is Fourier’s Law? Mathematically express Fourier’s Law defining all the terms used within it. What is thermal conductivity? Compare the values of thermal conductivity of metals, insulating materials and gases. Why does Fourier's law have a minus sign? According to the Fourier’s law, the rate of heat transfer through a material is directly proportional to the negative temperature gradient and the area of the material right angled to the temperature gradient. Therefore,  By using the constant k, the Fourier equation is formed as  Here  is the rate of heat transfer, A is the area of cross section of material through which the heat transfer takes place ,  is the temperature gradient over the material and the introduced constant k termed as the thermal conductivity of the material. (Thirumaleshwar, 2006, pp. 13-14) Thermal conductivity of a material reflects the heat conducting capacity of a material, and is measured in w/m-K. Thermal conductivity values of metals, insulating materials and gases are compared in the below table. Type Of Material Thermal Conductivity Value Metals 52 ~ 415 w/m-K Insulating materials 0.035 ~ 0.173 w/m-K Gases 0.0069 ~ 0.173 w/m-K Heat tends to flow from a hot surface to the cooler surface of a body. So it flows in the direction of decreasing temperature gradient, and this is why we take a negative sign in the formula (Adkins, 1987, p. 31). 4) Explain the Stefen-Boltzman Law. What is emissivity? What are the range of values for the emissivity of a surface? Define the terms “black surface” and “grey surface”. What role does the view factor play in determining the rate of heat transfer? What is a blackbody? Radiation of heat unlike conduction and convection doesn’t require any material medium to propagate. Stefan Boltzmann law gives a relationship between the maximum temperature of emitting surface and the surface temperature of the emitting surface. It states that the amount of heat radiated by a surface is proportional to the fourth power of its absolute temperature. It is given by the equation Qe(max) = ATS4 Where Qe(max) is the amount of radiation emitted per second, A is the area of the radiating surface, TS is the absolute temperature of the surface in Kelvin and  is Stefan’s constant. It is calculated to be 5.67 × 10-8 W/m2-k4 . The above equation applies when the body is a perfect radiator, i.e. a black body. However in practice since we don’t find ideal radiators, emissivity (ε) was defined to be the ratio of the energy emitted by a material to the energy radiated by a black body, having the same temperature. It is expressed as ε = Since emissivity is a ratio, it is a dimensionless quantity that depends on the material and the temperature at which it is measured at. Values of ε range from 0 to 1, a perfect black body would have an emissivity of 1 Hence, the Stefan-Boltzmann’s law for non-ideal radiators becomes: QeATS4 A black surface has a perfect emissivity of 1, as it a perfect radiator. In contrast a grey body has emissivity values between 0 and 1. View factor is defined as the fraction of radiation that radiates from surface 1 and strikes surface 2. It depends upon the geometry and relative positions of the surfaces and determines the amount of heat that is transferred when radiation takes place. A black body is a perfect absorber and a radiator. Any light that is incident upon it is completely absorbed and hence there is no reflected light from the surface of a black body. Because of this a black body appears to be absolutely black (Kjell, 2002, pp. 103-104). 5) Define heat of combustion, heat release rate and combustion reaction giving appropriate equations. Explain the different types of combustion and definitions of the following: Specific heat capacity, latent heat, calorimetry, combustion temperature and chemical equilibrium. Heat of combustion is defined as the total heat emitted when a unit quantity of a substance is combusted to form stable end products. It is measured in kJ/g or kJ/mole. The amount of heat released per unit time, when a substance is combusted is known as hear release rate, and is having the dimensions of Energy/ Time. So it is dimensionally identical to power and is measured in kilowatts(kW) or megawatts(mW). Combustion is defined as the chemical reaction between a fuel and an oxidant, which produces energy in the form of heat and light. Common combustion reactions include flame and explosions. Specific heat capacity is defined as the amount of heat required per unit mass of the substance that is required to raise the temperature by one degree centigrade. This should not be measured near temperatures involving phase change, as heat added or removed then would not change the temperature. Latent heat is the heat involved during phase changes of a substance. It is the heat that is absorbed or released during state changes without a change in temperature. Calorimetry is the measurement of heat that is absorbed or released during a chemical reaction, change in state or when substances dissolve to form solutions. Combustion temperature is the lowest temperature at which a substance ignites, and continues combusting when heated in air. It is also known as ignition point. Chemical equilibrium is the state of a reaction mixture when the rate of forward reaction is equal to the rate of the backward reaction. The reaction doesn’t proceed in any direction and the amounts of products and reactants remain constant. 6) Define the following terms: a. Heat capacity, Heat capacity is defined as the amount of heat energy required to increase the temperature of a given substance by 1˚C. It is measured in joules per kelvin. b. Specific heat, Specific heat is defined as the amount of heat required to raise the temperature of one kilogram of a substance by 1˚C. It is an intensive property and hence independent on the amount of the substance present. It is measured in joules per kilogram kelvin. c. Isothermal, isobaric, and isochoric processes Isothermal Process: A process that takes place at constant temperature is known as isothermal process. Isobaric Process: A process that takes place at constant pressure is known as isobaric process. Isochoric Process: A process in which the volume of the undergoing system remains constant is known as isochoric process (Vishnoi & Shukla, 2010, p. 8). 7) 61.6 ml of milk at 18.6 °C are added to 455.5 ml of coffee at 90.2 °C. What is the final temperature in degrees Celsius of this liquid mixture when thermal equilibrium is reached? Assume coffee has the same properties as pure water. The average density of milk is 1032 kg/m3. The specific heat of milk is 1.97 J/g °C. To reach the equilibrium the amount of heat lost by coffee is equal to the amount of heat gained by the milk. So, qmilk = -qcoffee , It implies that mmilk X cmilk X ∆Tmilk = - mcoffee X ccoffee X ∆Tcoffee We find the respective masses of milk and coffe by using their densities and volumes. Hence, mass of milk = 61.6 X 1.032 = 63.57g Similarly, mass of coffee= 455.5 X 1 = 455.5g Let te be the temperature at equilibrium Substituting in the above equation we have 63.57 X 1.97 X (18.6 - te ) =- 455.5 X 4.18 X ( 90.2 - te) We obtain the temperature at equilibrium te = 85.77 °C. 8) Consider two closed systems A and B. System A contains 3000 kJ of thermal energy at 20°C, whereas system B contains 200 kJ of thermal energy at 50°C. Now the systems are brought into contact with each other. Determine the direction of any heat transfer between the two systems. Heat always flows from hot body to cold body regardless of the thermal energy of the system. Hence heat flows from system B to system A due to the negative temperature gradient existing between them. 9) Explain Newton’s Law of cooling and give the mathematical equation defining all the terms used. How is natural convection different from forced convection? According to Newton’s law of cooling the rate of temperature change of a body is directly proportional to the difference between its temperature and the surrounding temperature. A hotter body radiates heat to its surroundings at a rate given by  (Tenv – T0) Where  is the net radiation heat loss, A = Surface area, T0 = Temperature of the object's surface ,Tenv = temperature of the environment and h is the heat transfer coefficient. If the environment is cooler Tenv < T0, and  is negative. This is true as a hotter body always radiates and thus loses heat to its cooler surroundings. h the coefficient of convective heat transfer can be defined as the heat transfer rate, in respect of unit surface area of a fluid and a solid surface, per unit temperature difference. As a mathematical equation, it is given by h =  , and its SI units are W/m2-K. Natural convection and Forced convection are two methods of convection, which differ in respect of their causes. Natural convection occurs when medium particles move on their own accord and transfer heat. Forced convection occurs when an external agent brings about the motion in medium particles. For example, a fan cooling the room is an example of forced convection (Coulson, et al., 1999, p. 414). 10) Aluminium has a specific heat of 0.902 J/goC. How much heat is lost when a piece of aluminium with a mass of 28.984 g cools from a temperature of 615.0oC to a temperature of 122.0oC? We know that q=mc∆T Here m=23.984g, c=0.902J/goC and ∆T = (615 – 122)= 493oC. Hence, q=23.984 X 0.902 X 493 = 10665.35 J heat is lost. List of References Adkins, C. J., 1987. An introduction to thermal physics. Cambridge University Press. Coulson, J. M., Richardson, J. F. & Backhurst, J. R., 1999. Coulson & Richardson's Chemical Engineering: Fluid flow, heat transfer, and mass transfe. 6th, ed. Elsevier. Kjell, J. G., 2002. Optical metrology. John Wiley and Son. Thirumaleshwar, M., 2006. Fundamentals of Heat and Mass Transfer. Pearson Education. Vishnoi, N. K. & Shukla, R. J., 2010. Textbook of Physical Chemistry. Ane Books Pvt Ltd. Read More

This should not be measured near temperatures involving phase change, as heat added or removed then would not change the temperature. Latent heat is the heat involved during phase changes of a substance. It is the heat that is absorbed or released during state changes without a change in temperature. Calorimetry is the measurement of heat that is absorbed or released during a chemical reaction, change in state or when substances dissolve to form solutions. Combustion temperature is the lowest temperature at which a substance ignites, and continues combusting when heated in air.

It is also known as ignition point. Chemical equilibrium is the state of a reaction mixture when the rate of forward reaction is equal to the rate of the backward reaction. The reaction doesn’t proceed in any direction and the amounts of products and reactants remain constant. 6) Define the following terms: a. Heat capacity, Heat capacity is defined as the amount of heat energy required to increase the temperature of a given substance by 1˚C. It is measured in joules per kelvin. b.

Specific heat, Specific heat is defined as the amount of heat required to raise the temperature of one kilogram of a substance by 1˚C. It is an intensive property and hence independent on the amount of the substance present. It is measured in joules per kilogram kelvin. c. Isothermal, isobaric, and isochoric processes Isothermal Process: A process that takes place at constant temperature is known as isothermal process. Isobaric Process: A process that takes place at constant pressure is known as isobaric process.

Isochoric Process: A process in which the volume of the undergoing system remains constant is known as isochoric process (Vishnoi & Shukla, 2010, p. 8). 7) 61.6 ml of milk at 18.6 °C are added to 455.5 ml of coffee at 90.2 °C. What is the final temperature in degrees Celsius of this liquid mixture when thermal equilibrium is reached? Assume coffee has the same properties as pure water. The average density of milk is 1032 kg/m3. The specific heat of milk is 1.97 J/g °C. To reach the equilibrium the amount of heat lost by coffee is equal to the amount of heat gained by the milk.

So, qmilk = -qcoffee , It implies that mmilk X cmilk X ∆Tmilk = - mcoffee X ccoffee X ∆Tcoffee We find the respective masses of milk and coffe by using their densities and volumes. Hence, mass of milk = 61.6 X 1.032 = 63.57g Similarly, mass of coffee= 455.5 X 1 = 455.5g Let te be the temperature at equilibrium Substituting in the above equation we have 63.57 X 1.97 X (18.6 - te ) =- 455.5 X 4.18 X ( 90.2 - te) We obtain the temperature at equilibrium te = 85.77 °C. 8) Consider two closed systems A and B.

System A contains 3000 kJ of thermal energy at 20°C, whereas system B contains 200 kJ of thermal energy at 50°C. Now the systems are brought into contact with each other. Determine the direction of any heat transfer between the two systems. Heat always flows from hot body to cold body regardless of the thermal energy of the system. Hence heat flows from system B to system A due to the negative temperature gradient existing between them. 9) Explain Newton’s Law of cooling and give the mathematical equation defining all the terms used.

How is natural convection different from forced convection? According to Newton’s law of cooling the rate of temperature change of a body is directly proportional to the difference between its temperature and the surrounding temperature. A hotter body radiates heat to its surroundings at a rate given by  (Tenv – T0) Where  is the net radiation heat loss, A = Surface area, T0 = Temperature of the object's surface ,Tenv = temperature of the environment and h is the heat transfer coefficient.

If the environment is cooler Tenv < T0, and  is negative. This is true as a hotter body always radiates and thus loses heat to its cooler surroundings. h the coefficient of convective heat transfer can be defined as the heat transfer rate, in respect of unit surface area of a fluid and a solid surface, per unit temperature difference.

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