Homework #7 (chapter 23): The First Law of Thermodynamics

Exercise 2:

H = k A DT/Dx = 0.74 W*m^-1*K^-1*6.2m*3.8m*44K*(0.32m)^-1 = 2.4 kW

Exercise 16:

a) If all potential energy of a representative chunk of water of mass Dm were converted to kinetic energy as it falls and then to thermal energy as heat, then

Dm g h = Dm c DT è DT = gh/c = 9.8 m/s² * 49.9 m / (4190 J*kg^-1*K^-1) = 0.12 K

b) Energy losses due to evaporative cooling, kinetic energy lost to splashing etc.

Exercise 23:

a) c = Q * m^-1 * DT^-1 = 320 J * (0.0371 kg)^-1 * (15.9 K)^-1 = 542 J*kg^-1*K^-1

b) n = m * M^-1 = 37.1 g * (51.4 g/mol)^-1 = 0.722 mol

c) c^mol = c * M = 542 J*kg^-1*K^-1 * (0.0514 kg/mol) = 27.9 J*mol^-1*K^-1

Problem 18:

1) adiabatic compression è gas heats up

2) heat extracted to cool gas to 0°C

3) isothermal expansion with heat added to keep gas at 0°C

Melting 122 g of ice at 0°C requires

Q = m L = 0.122 kg * 333 kJ * kg^-1 = 40.6 kJ

DE_int^cycle = 0 = W^cycle + Q^cycle with Q^cycle < 0 to melt the ice

è W^cycle = -Q^cycle = 40.6 kJ

Problem 19:

Monoatomic ideal gas: DE_int = 3/2 nRT = 3/2 RT

a) W^AB = 0 as DV = 0

DE_int^AB = 3/2 nR(T_B-T_A) = 3/2 * 1 mol * 8.314 J*K^-1*mol^-1 * 300 K = 3.74 kJ

Q^AB = DE_int^AB – W^AB = 3.74 kJ –0 = 3.74 kJ

Q^BC = 0

DE_int^BC = 3/2 nR(T_C-T_B) = 3/2 * 1 mol * 8.314 J*K^-1*mol^-1 (-145K) = -1.81 kJ

W^BC = DE_int^BC – Q^BC = -1.81 kJ – 0 = -1.81 kJ

DE_int^CA = 3/2 nR(T_A-T_C) = 3/2 * 1 mol * 8.314 J*K^-1*mol^-1 (-155K) = -1.93 kJ

Q^CA = n c_p (T_A – T_C) = 5/2 nR(T_A – T_C) = 5/2 * 1 mol * 8.314 J*K^-1*mol^-1 (-155K) = -3.22 kJ

W^CA = DE_int^CA – Q^CA = -1.93 kJ – (-3.22 kJ) = 1.29 kJ

b) Since n = 1 mol, pV = RT

p_A = 1.013 * 10⁵ Pa

V_A = R * T_A * p_A^-1 = 8.314 J*K^-1*mol^-1 * 300 K * (1.013 * 10⁵ Pa)^-1 = 0.0247 m³

V_B = V_A = 0.0247 m³

p_B = R * T_B * V_B^-1 = 8.314 J*K^-1*mol^-1 * 600 K * (0.0247 m³)^-1 = 2.02 * 10⁵ Pa

p_C = p_A = 1.013 * 10⁵ Pa

V_C = R * T_C * p_C^-1 = 8.314 J*K^-1*mol^-1 * 455 K * (1.013 * 10⁵ Pa)^-1 = 0.0374 m³