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Consider the adhering to statements:A. The entropy of an isolated system never before decreases.B. Heat never flows spontaneously from cold to warm.C. The full thermal energy of an isolated mechanism is continuous.Which of these express the second regulation of thermodynamics?
A heat engine is designed to perform work. This is feasible just if specific relationships between the heats and also temperatures at the input and output hold true.If Q_h is the magnitude of warm soaked up, and also Q_c is the magnitude of warmth liberated, what need to be true of the relationships between Q_h and Q_c and also T_h and also T_c for a warm engine to do work?
Given heat engines:A. P_in=1000W, T_h = 600K, P_out=600W, T_c=300K, Output Power = 400 WB. P_in=1000W, T_h = 600K, P_out=0W, T_c=300K, Output Power = 1000 WC. P_in=1000W, T_h = 600K, P_out=600W, T_c=200K, Output Power = 600 WD. P_in=1000W, T_h = 500K, P_out=500W, T_c=300K, Output Power = 500 WE. P_in=1000W, T_h = 500K, P_out=550W, T_c=250K, Output Power = 450 WF. P_in=1000W, T_h = 600K, P_out=250W, T_c=200K, Output Power = 650 WWhich of these violates the first legislation of thermodynamics?Of those remaining, which violate the second regulation of thermodynamics?Which of the staying deindicators has actually the greatest thermal efficiency?
C, F∆E_th = W + QW is Output Power, Q is power in, ∆E_th is power outB,DFind e_max using Carnot"s 1 - (T_c / T_h), then find e making use of W_out / Q_h. if e > e_max, the second regulation has been violated.E.45 compared to A"s .4
calculates max efficiencye_max = 1 - (T_c / T_h)where T is the Kelvin temperature of a cold and also a hot reservoir
While keeping your food cold, your refrigerator transfers energy from the inside to the surroundings. Hence, thermal power goes from a chillier object to a warmer one.Does this violate the second legislation of thermodynamics? Why or why not?
The second regulation of thermodynamics uses in this instance, yet it is not violated bereason the power did not spontaneously go from cold to hot.
coeffective of performancecalculation relies on whether the heat pump is used for cooling (refrigerator) or for heating (geothermal heating system)
COP = what you acquire / what you paidfor cooling, what you acquire = Q_cfor heating, what you gain = Q_heither means, what you paid is W_inCOP_max = temp you want / (temp of hot reservoir - temp of cool reservoir)for cooling, temp you desire is T_cfor heating, temp you want is T_h
Calculate the coeffective of performance for a refrigerator that takes in 40 J of occupational and also exhausts 100 J of warm.
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