Chapter 20
The First Law of Thermodynamics
EXAMPLES
Chapter 20. Heat and the First Law of
Thermodynamics:
Examples
Example 20.1 A very Hot Frying Pan
Accidentally an empty iron frying pan gets very hot on the stove
(~300oC). What happen when you dunk it into a few inches of cool water
in the bottom of the sink?
Assume: Mwater ≈ Mpan
Water warms up (~10oC to 20oC)
Water does not come close to boiling.
 Since: Mwater ≈ Mpan & cwater ~ 10ciron  from
c
Q
Q
 T 
m T
cm
heat Q leaving the pan enters the water, and the iron’s pan ΔT will be
10 times greater than ΔT of the water.
If you let a few drops of water fall onto the hot pan, that very small mass
of water will sizzle and boil away since:
Mpan ~ 100Mwater
Example 20.2 Cooling a Hot Ingot
A 0.05-kg ingot of metal is heated to 200oC and then dropped into a
beaker containing 0.4-kg of water initially at 20oC. If the final
equilibrium temperature of the mixed system is 22.4oC, find specific
heat (c) of the metal.
cs 
mwcw T f  Tw 
ms Ts  T f 
cs  453 J/kg  C
(0.4 kg)(4186 J/kg  o C)(22.4 o C  20.0 C)


(0.05kg)(200.0 C  22.4 C )
Calorimetry Problem-Solving Strategy
Units of measure must be consistent
 For example, if your value of c is in J/kg.oC, then your mass must be
in kg, the temperatures in oC and energies in J
Transfers of energy are given by Q =mc ΔT only when no phase change
occurs
If there is a phase change, use Q = mL
 Be sure to select the correct sign for all energy transfers
Remember to use Qcold = –Qhot
 The ΔT is always Tf - Ti
Example 20.3 Cooling the Steam
What mass of steam initially at 130oC is needed to warm 200-g of water in
a 100-g glass container from 20oC to 50.0oC
Steam loses energy in 3 stages:
1st stage: Steam  100oC , DT = –30oC


 

Q1  ms cs T  ms 2.01x103 J / kg o C 30o C   6.03x104 J / kg ms
2nd stage: Steam  water


Q2  ms Lv   2.26 x106 J / kg ms
3rd stage: Water  50oC , DT = –50oC





Q3  ms cw T  ms 4.19 x103 J / kg o C 50o C   2.09 x105 J / kg ms
Example 20.3 Cooling the Steam, final
Adding the energy transfer in the 3 stages:


Qhot  Q1  Q2  Q3   2.53x106 J / kg ms
Water and Glass increasing temperature: DT = 30oC


Qcold  mwcw T  mg cg T   0.200kg  4.19 x103 J / kg o C 30o C




  0.100kg  837 J / kg o C 30o C 
Qcold  2.77 x104 J
Final: Qcold = –Qhot


2.77 10 4 J    2.53 106 J / kg)ms  ms  1.110 2 kg  11g
Example 20.4 An Isothermal Expansion
A 1.0 mol of an ideal gas is kept at 0oC during an expansion from 3.0 L to
10.0 L.
(A). Find the work done on the gas during the expansion. Using equation
20.13
V 
 3.0L 
3
W  nRT ln  1   1mol 8.31J / mol  K  273K  ln 


2.7
x
10
J

 10.0L 
 V2 
(B). How much energy transfer by heat occurs with the surroundings in
this process?
 From the first law: ΔEint = Q + W
Ein  0  0  Q  W  Q  W  2.7 x103 J
Example 20.4 An Isothermal Expansion, final
(C). Find the work done on the gas during the expansion.
 Work done in an isobaric process is: W = P (Vf – Vi)
where Vf =10.0L and Vi =3.0L (reverse of part A)
 nRTi 
W   P V f  Vi   
 V f  Vi  
 Vi 
1.0mol 8.31J / mol  K  273K  3.0  10.0 x103 m3 
W 


10.0 x103 m3
W  1.6 x103 J
Example 20.5 Quick Quiz 20.5
Two rods from different materials with the same length and diameter
are connecting two regions of different temperatures. In which case is
the rate of energy transfer by heat (P ) larger
Series
Parallel
 T T 
PS  kA  h c 
 2L 
Therefore PP = 4 PS
 T T 
PP  k 2 A  h c 
 L 
: In parallel, the rods present a larger
area A and a smaller length L through which the energy can transfer.
Material for the 2nd Midterm
Material from the book to Study!!!
Objective Questions: 6-12
Conceptual Questions: 3-5
Problems: 6-12-16-22-23-24-25-2630-36-49