Answers to homework problems

Chapter 11

14) E = m*c*deltaT, where E is an energy, m is mass and deltaT is the
    change in temperature =>
    deltaT = E/(m*c) = 38.22 degrees => Tfinal = 58.3 degrees centigrade.

16) E = P*t, where P is a power and t is the time, then
    E = m*c*(Tfinal -Tinitial), where c is a specific heat of the water
    m is a mass of the water, Tinitial is the initial temperature of the water
    and Tfinal is a final temperature of the water.
    P*t = m*c*(Tfinal - Tinitial) =>
    t = m*c*(Tfinal - Tinitial)/P = 
      = 1kg * 4200J/kg * (100-20) / 800W = 420s = 7 min

20) Consider a mass of water, m, falling from the height H = 75 meters.
    The change in gravitational energy is deltaP = m*g*H 
    The energy needed to raise the water temperature by a  deltaT is:
       deltaP = m*c*deltaT =>
    m*c*deltaT = m*g*H => deltaT = g*H/c = 9.8 * 75/4200 = 0.175 degrees.

21) I find that the crucial information is missing from the book and has 
    to be found in tables of material in libraries. This is formation is
    the density of aluminum: rho(Al) = 2.7  g/cm^3
    the density of copper:   rho(Cu) = 8.96 g/cm^3

    The solution is as follows:
    Change of the energy of the aluminum block:
      dE = V * rho(Al) * c(Al) * dT1
     Here  V is volume, rho(Al) is density, c(Al) is specific heat of 
    the aluminum cube, and dT1 is the change in temperature
    dT1 = 100 - 20  (remember room temperature =20 degrees C)
    The same amount of energy is added to the copper block of similar 
    dimensions: 
      dE = V * rho(Cu) * c(Cu) * dT2 ,
    with c(Cu) - specific heat of copper and dT2 - temperature raise
    from the room temperature.
    The two changes of energy must be equal
     V * rho(Al) * c(Al) * dT1 = V * rho(Cu) * c(Cu) * dT2 
       => dT2 = rho1 * c1 * dT1 / (rho2 * c2)  = 57 degrees
    The final temperature is T = room temperature +dT2 = 77 degrees C
              

Chapter 12

12) Energy W = dU + dQ, where dQ is work done by the gas and dU is the
    change of the internal energy of the gas. Since the container is rigid
    then process is ISOMETRIC since volume does not change and dQ = PdV = 0.
    Therefore  W = dU: 3 kcal = 3 * 4186 J = 12558 J
    
14) Adiabatically means Q=0 (no heat exchange)
           dU + W = 0
   a) Since W is positive (gas performes work) then dU is negative 
      and the temperature decreases. This is how your refrigerator works !
   b) adiabatic means no heat transfer 
   c) the lowering of temperature is done at a cost of internal energy 
      dU = -500 J

22) Work done by gas in p-V diagram is simply the area under the 
      initial and final points in the diagram. W= p * delta p.
      W(1-2) = 0  (constant volume = no work done)
      W(2-3) = 0.5 * 10^5 * (0.75 - 0.25) = 2.5 10^4 J
      W(3-4) = 0  (constant volume = no work done)
      W(4-5) = 1.0 * 10^5 * (1.00- 0.75) = 2.5 10^4 J

24) The same argument as in problem 22: W = area under the curve
      
          W = {(2*10^5 + 5*10^5)/2}*(1-0.5) = 1.75*10^5J