FUND OF ENG THERMODYN(LLF)+WP NEXT GEN
9th Edition
ISBN: 9781119840602
Author: MORAN
Publisher: WILEY
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A firearm can be modeled as a kind of heat engine, where the projectile acts as a piston that separates from the rest of the system during
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thermal efficiency of the "engine"). The other 98.9% can be approximated as being entirely absorbed by the barrel, which increases in
temperature uniformly for a short time before this energy is dissipated into the surroundings. What is this temperature increase (in °C)?
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°℃
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Chapter 2 Solutions
FUND OF ENG THERMODYN(LLF)+WP NEXT GEN
Ch. 2 - Prob. 2.1ECh. 2 - Prob. 2.2ECh. 2 - Prob. 2.3ECh. 2 - Prob. 2.4ECh. 2 - Prob. 2.5ECh. 2 - Prob. 2.6ECh. 2 - Prob. 2.7ECh. 2 - Prob. 2.8ECh. 2 - Prob. 2.9ECh. 2 - Prob. 2.10E
Ch. 2 - Prob. 2.11ECh. 2 - Prob. 2.12ECh. 2 - Prob. 2.13ECh. 2 - Prob. 2.14ECh. 2 - Prob. 2.15ECh. 2 - Prob. 2.16ECh. 2 - Prob. 2.17ECh. 2 - Prob. 2.1CUCh. 2 - Prob. 2.2CUCh. 2 - Prob. 2.3CUCh. 2 - Prob. 2.4CUCh. 2 - Prob. 2.5CUCh. 2 - Prob. 2.6CUCh. 2 - Prob. 2.7CUCh. 2 - Prob. 2.8CUCh. 2 - Prob. 2.9CUCh. 2 - Prob. 2.10CUCh. 2 - Prob. 2.11CUCh. 2 - Prob. 2.12CUCh. 2 - Prob. 2.13CUCh. 2 - Prob. 2.14CUCh. 2 - Prob. 2.15CUCh. 2 - Prob. 2.16CUCh. 2 - Prob. 2.17CUCh. 2 - Prob. 2.18CUCh. 2 - Prob. 2.19CUCh. 2 - Prob. 2.20CUCh. 2 - Prob. 2.21CUCh. 2 - Prob. 2.22CUCh. 2 - Prob. 2.23CUCh. 2 - Prob. 2.24CUCh. 2 - Prob. 2.25CUCh. 2 - Prob. 2.26CUCh. 2 - Prob. 2.27CUCh. 2 - Prob. 2.28CUCh. 2 - Prob. 2.29CUCh. 2 - Prob. 2.30CUCh. 2 - Prob. 2.31CUCh. 2 - Prob. 2.32CUCh. 2 - Prob. 2.33CUCh. 2 - Prob. 2.34CUCh. 2 - Prob. 2.35CUCh. 2 - Prob. 2.36CUCh. 2 - Prob. 2.37CUCh. 2 - Prob. 2.38CUCh. 2 - Prob. 2.39CUCh. 2 - Prob. 2.40CUCh. 2 - Prob. 2.41CUCh. 2 - Prob. 2.42CUCh. 2 - Prob. 2.43CUCh. 2 - Prob. 2.44CUCh. 2 - Prob. 2.45CUCh. 2 - Prob. 2.46CUCh. 2 - Prob. 2.47CUCh. 2 - Prob. 2.48CUCh. 2 - Prob. 2.49CUCh. 2 - Prob. 2.50CUCh. 2 - Prob. 2.51CUCh. 2 - Prob. 2.52CUCh. 2 - Prob. 2.53CUCh. 2 - Prob. 2.54CUCh. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10PCh. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26PCh. 2 - Prob. 2.27PCh. 2 - Prob. 2.28PCh. 2 - Prob. 2.29PCh. 2 - Prob. 2.30PCh. 2 - Prob. 2.31PCh. 2 - Prob. 2.32PCh. 2 - Prob. 2.33PCh. 2 - Prob. 2.34PCh. 2 - Prob. 2.35PCh. 2 - Prob. 2.36PCh. 2 - Prob. 2.37PCh. 2 - Prob. 2.38PCh. 2 - Prob. 2.39PCh. 2 - Prob. 2.40PCh. 2 - Prob. 2.41PCh. 2 - Prob. 2.42PCh. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Prob. 2.45PCh. 2 - Prob. 2.46PCh. 2 - Prob. 2.47PCh. 2 - Prob. 2.48PCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Prob. 2.55PCh. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Prob. 2.59PCh. 2 - Prob. 2.60PCh. 2 - Prob. 2.62PCh. 2 - Prob. 2.63PCh. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. 2.67PCh. 2 - Prob. 2.68PCh. 2 - Prob. 2.69PCh. 2 - Prob. 2.70PCh. 2 - Prob. 2.71P
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- For a refrigerator or air conditioner, the coefficient of performance K (often denoted as COP) is the ratio of cooling output Qc to the required electrical energy input W, both in joules. The coefficient of performance is also expressed as a ratio of powers, |Qc\/t K |W\/t where |Qc|/t is the cooling power and W/t is the electrical power input to the device, both in watts. The energy efficiency ratio (EER) is the same quantity expressed in units of Btu for Qc and W h for W|. Part A Derive a general relationship that expresses EER in terms of K. Express your answer in terms of K. EER = Submit Request Answerarrow_forwardEntropy can be defined as a level of disorder. This disorder in not necessarily related to thermodynamics only. Discuss in your own words your understanding of the concept of entropy with respect to non-engineering applications of our life with the help of two examples. Discuss, in your own words, why is important to define and understand the concept of entropy.arrow_forwardA thermodynamics steady flow system receives 4.56 kg/min of a fluid where the pressure is 137.9 Kpa specific volume of 0.0388 m3/kg, velocity of 122m/s and internal energy of 1,160 Kj/Kg. the fluid leaves the system at a boundary where pressure is 551.6Kpa , specific volume of 0.193 m3/kg , velocity of 183 m/s and internal energy of 52.8 Kj/kg. during the passage through the system, the fluid receives 3,000 Joules/s of heat. Determine the steady flow work in. Kj/min and kj/sarrow_forward
- The molar heat capacities of substances varies with temperature. The general function for determining the molar heat capacity is given below; Cp m = (a + bT + cT 2 )R . In case of a gas where a = 3.245, b = 7.108 x10^ -4 K ^-1 , and c = -4.06 x10 ^-8 K^ -2 for temperatures in the range of 800 Kelvins to 1,500 Kelvins. What is the change in the enthalpy (in KiloJoules per Kelvin per mole) for two moles of this gas at 1,500 kelvins? NOTE: Express answer in THREE SIGNIFICANT FIGURES.arrow_forwardThe molar heat capacities of substances varies with temperature. The general function for determining the molar heat capacity is given below; Cp = ( a + b T + c T 2 )R . In case of a gas where a = 3.245, b = 7.108 x10 ^-4 K ^-1 , and c = -4.06 x10^ -8 K ^-2 for temperatures in the range of 300 Kelvins to 1,500 Kelvins. What is the heat capacity (in Joules per Kelvin per mole) of this gas at 1,500 kelvins? NOTE: Express answer in THREE SIGNIFICANT FIGURES.arrow_forwardA food product containing 80% moisture content is being frozen. Estimate the specific heat of the product at -6 ° C when 82% of the water is frozen. The specific heat of the dry product is 2.5 kJ / (kg ° C). it is assumed that the specific heat of water at -10 ° C is the same as the specific heat of water at 0 ° C, and the specific heat of ice follows the function Cp es = 0.0062 Tbeku + 2.0649. Cp frozen products = AnswerkJ / kg ° C.arrow_forward
- Give at least five (5) applications of Thermodynamics in the Real World and discuss how it is applied.arrow_forwardA food product containing 75% moisture content is being frozen. Estimate the specific heat of the product at -10 ° C when 85% of the water is frozen. The specific heat of the dry product is 2 kJ / (kg ° C). it is assumed that the specific heat of water at -10 ° C is the same as the specific heat of water at 0 ° C, and the specific heat of ice follows the function Cp es = 0.0062 Tbeku + 2.0649. Cp of frozen product = .... kJ / kg ° Carrow_forwardShow by calculation that heat is not a state function. Use the first law of thermodynamics and internal energy as a function of temperature and volume....arrow_forward
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