# Phy 121 - Assignment 13 ```Phy 121 - Assignment 13
A. 1. a. Hold a piece of metal over a flame, or put an egg in a hot frying pan, or
… (There are thousands of possible answers.)
b. Put an ice/water mixture in a hot frying pan. (If kept well stirred, it stays at
0&deg;C until completely melted. As energy is added, the internal energy goes up,
but the molecules are not gaining kinetic energy, so the temperature stays the
same.)
2.
c) First law: Q = ΔEint + W
ΔEint = Q – W = 3.46 – 1.00 =
2.46 kJ
B. 1. a. One of many possible answers: Start a fire by rubbing two sticks
together. The work done by friction, not heat flow, raises a stick’s internal
energy and its temperature.
b. Ride in an elevator. Work is done on you as you are lifted, but it increases
an external form of energy, having no effect on internal energy or temperature.
2.
C. 1. a. -100 J: From Q = ΔEint + W, ΔEint = Q - W. Q = 0 is the definition of
the word adiabatic. (No heat flow.) W, the work, is given to be 100 J. (The
answer is negative because Eint is decreasing: As the gas expands, it gives
energy to the outside world, perhaps turning the wheels of a car.)
b. 0 J: For a gas, Eint is proportional to temperature. ΔT = 0 is what the word
isothermal means.
2.
D. 1. No. A refrigerator works by pumping heat from its inside to its outside.
Having the door open just lets it flow back in again. That doesn't remove heat
from the room.
2. b) For the Carnot cycle,
(Remember
temperatures must be in kelvin.)
For any kind of engine,
, so W = (e)(Qin) = (.3824)(1200 J) = 459 J
a) Qin = W + Qout
1200 = 459 + Qout  Qout =
741 J
E.1. The decrease is not spontaneous. All living things take in energy from food
or light.
2. b) Qin = W + Qout
1600 J = W + 1000 J  W =
600 J
a)
=
c) Power =
.375
(or 37.5 %)
= 2000 W
F. 1. You need to say how friction and some other force cancel each other,
giving a total of zero. For example, pull a chair across the floor with a force of
10 N while 10 N of friction pulls the other way. Since the net force is zero, the
chair maintains a constant velocity.
2. ∑τ = 0
(3 m)(sin 90&deg;)(F2) – (2 m)(sin 35&deg;)(300 N) = 0
3F2 – 344.1 = 0
3F2 = 344.1
F2 =
=
115 N
```