# Chapter 11:Thermal Properties of Matter¶

## Ex11.1:pg-303¶

In [1]:
  import math   #Example 11_1

#To find out how much heat is required to change the temperature
#With 400 Grams of water
m=400     #Units in gm
q=c*m*t      #Units in Cal
print "The heat required for 400 gm of water is Q=",round(q)," Cal\n"
#With 400 grams of copper
m=400     #Units in gm
q=c*m*t      #Units in Cal
print "The heat required for 400 gm of copper is Q=",round(q)," Cal\n"

The heat required for 400 gm of water is Q= 2000.0  Cal

The heat required for 400 gm of copper is Q= -186.0  Cal



## Ex11.2:pg-303¶

In [2]:
  import math   #Example 11_2

#To findout how much water is released
#When it crystallizes
m=50      #Units in gm
h=80      #Units in Cal/gm
q=m*h      #Units in Cal
print "When it crystallizes heat required is Q=",round(q)," Cal\n"
#When it Condenses
m=50      #Units in gm
h=539      #Units in Cal/gm
q=m*h      #Units in Cal
print "When it condenses heat required is Q=",round(q)," Cal\n"
#In textbook answer is printed wrong as Q=27000 cal but the correct answer is Q=26950 Cal

When it crystallizes heat required is Q= 4000.0  Cal

When it condenses heat required is Q= 26950.0  Cal



## Ex11.3:pg-304¶

In [3]:
  import math   #Example 11_3

#To findout the amount of Ice that has to be added
m=200      #Units in gm
change=m*c*(tf-to)       #units in Cal
Hf=80      #Units in Cal/gm
change1=Hf+c*(tf-to)       #Units in Cal/gm
M=change/-(change1)
print "The amount of ice that has to be added is M=",round(M,1)," gm"

The amount of ice that has to be added is M= 54.0  gm


## Ex11.4:pg-305¶

In [5]:
  import math   #Example 11_4

#To findout the specific heat capacity of the metal
m=400        #Units in gm
oil=m*c*(tf-to)         #units in cal
m1=80        #Units in gm
cm=m1*(tf-to)         #units in in terms of cm and gm Centigrade
print "The specific heat of metal is Cm=",round(cmm,3)," cal/gm C"

The specific heat of metal is Cm= 0.216  cal/gm C


## Ex11.5:pg-305¶

In [6]:
  import math   #Example 11_5

#To findout how long does the heater takes to heat
m=500        #Units in gm
m1=30         #Units in gm
hv=65       #Units in cal/gm
Hg=((m*c*(tf-to))+(m1*hv))*4.1808135         #units in Joules
delivered=70        #Units in Joule/Sec
t=Hg/delivered       #Units in sec
print "The time taken is t=",round(t)," sec"
#In textbook answer printed wrong as t=450 sec correct answer is t=448 sec

The time taken is t= 449.0  sec


## Ex11.6:pg-306¶

In [7]:
  import math   #Example 11_6

#To findout the rise in temperature
m=0.01     #Units in Kg
v=100         #Units in meters/sec
KE=(0.5*m*v**2)/4.1808135        #units in Cal
m=10      #units in gm
t=KE/(m*c)
print "the rise in temperature is DeltaT=",round(t,1)," C"

the rise in temperature is DeltaT= 38.6  C


## Ex11.8:pg-307¶

In [8]:
  import math   #Example 11_8

#To findout how much longer is at 35 degrees
dist=20.0   #Unis in meters
L=alpha*dist*t        #Units in meters
print "The slab is longer by=",round(L,3)," meters"

The slab is longer by= 0.01  meters


## Ex11.9:pg-308¶

In [9]:
  import math   #Example 11_9

#To findout how large a diameter  when the sheet is heated
dist=2         #Units in cm
L=dist*delta*t           #Units in cm
print "The new diameter of the hole is=",round(2+L,4)," cm"

The new diameter of the hole is= 2.0076  cm


## Ex11.10:pg-309¶

In [10]:
  import math   #Example 11_10

#To findout the change in benzene volume
v10=100.0         #Units in cm**3
v20=delta*t+v10       #Units in cm**3
V=v20*delta*t     #Units in cm**3
v30=V+v20     #Units in cm**3
print "The change in benzene volume is V30=",round(v30,3)," cm**3"
#In textbook the answer is printed wrng as V3=0102.5 cm**3 the correct answer is V3=101.253 cm**3

The change in benzene volume is V30= 101.253  cm**3


## Ex11.11:pg-309¶

In [11]:
  import math   #Example 11_11

#To findout how much ice melts each hour
s=30      #Units in cm
a=s*s*10**-4      #units in meter**2
k=0.032     #Units in W/K meter
t=25    #Units in K
l=0.040         #Units in meters
q_t=(6*k*((a*t)/l))/4.1808135      #Units in cal/sec
Q=3600*q_t      #Units in cal
qq=80      #Units in cal/gm
melted=Q/qq        #Units in gm
print "The ice melts by ",round(melted)," gm"

The ice melts by  116.0  gm


## Ex11.12:pg-310¶

In [12]:
  import math   #Example 11_12

#To compare the energy emitted per unit area of our body to with the same emissivity
t1=273+t1         #Units in K
t2=273+t2         #Units in K
tb_tc=(t1/t2)**4         #Units in terms of (Tb/Tc)**4
tb_tc=tb_tc*100         #In terms of percentage
print "The radiation defers by ",round(tb_tc-100)," percent"

#In textbook answer is printed wrong as 40% the correct answer is 34%

The radiation defers by  34.0  percent


## Ex11.13:pg-310¶

In [13]:
  import math   #Example 11_13

#To findout how much heat is lost through it
a=15     #Unis in meter**2
t=30.0      #Units in K
R=2.2   #Units in Meter**2 K/W
q_t=(a*t)/R      #Units in W
T=3600.0       #Units in sec
Q=q_t*T         #Units in J
print "The amount of heat lost is Q=",round(Q,1)," J"

The amount of heat lost is Q= 736363.6  J