University Physics (OpenStax) Vol 2, 2021
P2311 Homework 11 (Ch. 1 - Temperature)
Assigned:
Problems
Ch. 1 Problems: 43-45,49,52,54,61-63,70,72,85
43. Celsius.
Your Fahrenheit temperature is 102°F.
Yes, it is time to get treatment.
44. a) 293 K, b) 329.6 K c) 5750 K
45. a) Delta T_C = 22.2 C
b) Delta T_F = 9/5 Delta T_C
49. Use L - Lo = Lo alpha Delta T
L = Lo (1+ alpha Delta T) = 169.98 m
52. Use Delta L = gap
Delta L = L alpha_steel Delta T
= 10.0(12e-6)(30) = 3.6e-3 m = 3.6 mm
54. alpha_water = beta_water/3 = 70e-6 C^-1
Delta H = H alpha_water Delta T
Delta H = .070 m = 7 cm But this assumes that the
other two dimensions can expand freely. If they can't
(as in this problem), then Delta H = 3 x 7 = 21 cm,
in order for the volume change to obey Delta V = 3 alpha Delta T.
So Delta H = 21 cm.
61. Use Q = m c Delta T
Q=5.02e+8 J (key says "m=5.02e+8 J")
62. Use Q = m c_glass Delta T
Q = 3066 J
63. Use Tf-Ti = Q/mc
a) Tf = 21 C
b) Tf = 25 C
c) Tf = 29.3 C
d) Tf = 50 C
70. Use Q_water = -Q_steel
mc(Te-10) = -m_s c_s (Te-215)
Te = 16.45 C
72. Q = +L_f m with L_f = 79800 cal/kg for water
Q = 35.91 kcal
85. Use Q_cube = -Q_water
Check how much energy the liquid water can loose to the ice
before reaching 0 C: 58604 J. Since this is bigger than
Q_melt and Q_warm_ice, the ice will all melt and the equation is:
Q_warm + Q_melt + Q_warm_meltwater = - (Q_cool)
3135 + 16702 +209.3 T_f = 58604 - 1674.4 T_f
T_f = 20.58 C
Check
#49 & #70. -1/2 pt for any conceptual question missing.
-1/2 pt for 2-4 problems with no work
-1 pt for >4 problems with no work.