Chapter 5, Problem 5.14CU

According to Newton’s law of cooling, the rate of change of temperature is proportional to the temperature difference between the system and its surroundings:dT/dt= -α(T-Tsur)where Tsur is the temperature of the surroundings and α is a constant. (a) Integrate this equation with the initial condition that T = Ti at t = 0.(b) Given that the entropy varies with temperature according to S(T) − S(Ti) = C ln(T/Ti), where Ti is the initial temperature and C the heat capacity, deduce an expression entropy of the system at time t.

An ideal gas undergoes a process from state 1 ( the properties are T₁ = 300 K, p₁ = 100 kPa) to state 2 (the properties are T₂ = 600 K, p₂ = 500 kPa). The specific heats of the ideal gas are: c = 1 kJ/kg-K and c = 0.7 kJ/kg-K.. The change in specific entropy of the ideal gas to two decimal places)from state 1 to state 2 (in kJ/kg-K) is......

One-quarter Ibmol of oxygen gas (O2) undergoes a process from p1 = 20 Ib/in?, T1 = 500°R to p2 = 150 lb;/in?. For the process W = -500 Btu and Q = -127.5 Btu. Assume the oxygen behaves as an ideal gas. Determine T2, in °R, and the change in entropy, in Btu/°R. Step 1 Determine T2, in °R. T2 = °R Save for Later Attempts: 0 of 1 used Submit Answer Step 2 The parts of this question must be completed in order. This part will be available when you complete the part above.

Derive an expression for the change in entropy of the universe. 7

One-quarter Ibmol of oxygen gas (O₂) undergoes a process from p₁ = 20 lbf/in², T₁ = 500°R to p₂ = 150 lb/in². For the process W = -500 Btu and Q = -202.5 Btu. Assume the oxygen behaves as an ideal gas. Determine T2, in °R, and the change in entropy, in Btu/°R.

The entropy change of a system can be negative, but the entropy generation cannot.

One-quarter Ibmol of oxygen gas (O2) undergoes a process from p1 = 20 lb/in?, T1 = 500°R to p2 = 150 lb;/in?. For the process W = -500 Btu and Q = -202.5 Btu. Assume the oxygen behaves as an ideal gas. Determine T2, in °R, and the change in entropy, in Btu/°R.

find the changes in h  as appropriate. The initial state pressure is p1 = 0.5 MPa. the final state is 2.   a.  constant volume :                   v1 = 0.3 m3/kg,            p2 = 0.3 MPa; b. constant entropy :                    s1 = 6.3 kJ/kg K,        p2 = 0.15 MPa; c. constant volume :                    h1 = 2500 kJ/kg,          p2 = 0.2 MPa; d. constant enthalpy :                 s1 = 6.4 kJ/kg K,        p2 = 0.2 MPa;

A certain material has the Helmoltz free energy given as: 3 V F(T,V,n) = –nRT Ln|CTZ nRT + na Where a, b, and C are constants. Calculate the following quantities as a function of T, V, and n. 1- Entropy 2- Pressure 3- Internal energy 4- Chemical potential 5- Enthalpy 6- Gibbs free energy 7-Heat capacity at constant volume 8- Heat capacity at constant pressure