# Chapter 2 : Thermometry¶

## Example 2.1 pageno : 29¶

In [1]:
# Variables
li = 1.23;			#length of melting ice in mm
lf = 18.56;			#length of melting ice reading in pressure of 74.24cm of mercury in mm
l = 10.75;			#length of melting ice at which temperature to be calculated
mp = 0;			#melting point in deg.C
T = 50;			#temperature of melting ice at which length to be calculated in deg.C

# Calculations
sp = 100-(76-74.24)/(2.7);			#76cm of mercury steam point is 100 deg.C so at 74.24cm of mercury the steam point in deg.C
t = (l-li)*(sp-mp)/(lf-li);			#temperature at 10.75mm of melting ice in deg.C
lt = ((T*(lf-li))/(sp-mp))+li;			#length of ice at 50 deg.C

# Result
print 'the temperature of melting ice at 10.75mm of hg is %3.2f deg.C \
\nthe length of ice corresponding to 50 deg.C is %3.2f mm of mercury'%(t,lt)

the temperature of melting ice at 10.75mm of hg is 54.58 deg.C
the length of ice corresponding to 50 deg.C is 9.95 mm of mercury


## Example 2.2 pageno : 30¶

In [2]:
# Variables
p1 = 23.5;			#pressure when immersed in liquid air in cm
p2 = 75.;			#pressure when immersed in ice in cm
p3 = 102.4;			#pressure when immersed in steam in cm
T = 100.;			#boiling point of temperature in deg.C

# Calculations
t = (p1-p2)*T/(p3-p2);			#temperature of the liquid air in deg.C

# Result
print 'the temperature of liquid of air is %3.2f deg.C'%(t)

the temperature of liquid of air is -187.96 deg.C


## Example 2.3 pageno : 30¶

In [6]:
# Variables
t1 = 283.;			#temperature of bulb when pressure is h-2cm of hg in k
t2 = 546.;			#temperature of bulb when pressure is h-22cm of hg in k
h1 = 2.;			#differnce of mercury level at 283k in cm
h2 = 22.;			#differnce of mercury level at 546k in cm

# Calculations
h = ((h2*t1)+(h1*t2))/(t2-t1);			#height of the barometer in cm

# Result
print 'height of the barometer is %3.2f cm'%(h)

print "Note : Answer in book is wrong. Please calculate manually and check."

height of the barometer is 27.83 cm
Note : Answer in book is wrong. Please calculate manually and check.


## Example 2.4 pageno : 30¶

In [7]:
# Variables
p0 = 76.;			#pressure at 0 deg.C in cm of hg
p1 = 228.;			#pressure (76+152) at T deg.C in cm of hg
t0 = 273.;			#temperature of bulb in K

# Calculations
T = p1*t0/p0;			#temperature at 228 cm of hg pressure in K

# Result
print 'the temperature of bulb is %3.2f K'%(T)

the temperature of bulb is 819.00 K


## Example 2.5 page no : 30¶

In [13]:
# Variables
t1 = 0;			    #temperature in deg.C
t2 = 100;			#temperature in deg.C
t3 = 208;			#temperature in deg.C
r1 = 3.5;			#resistance in ohms
r2 = 5.2;			#resistance in ohms
r3 = 6.9;			#resistance in ohms
r4 = 9.4;			#resistance in ohms

# Calculations
tpt = (r3-r1)*100/(r2-r1);			#temperature in deg.C
d = round((t3-tpt)/(2.08*1.08),2);			#deflection
tp = round((r4-r1)*100/(r2-r1),2);			#temperature in deg.C
t6 = (3.5*(((tp/100)**2)-tp/100))+tp;			#temperature in deg.C
t7 = (3.5*(((t6/100)**2)-t6/100))+tp;			#temperature in deg.C
t8 = (3.5*(((t7/100)**2)-t7/100))+tp;			#temperature in deg.C
t9 = (3.5*(((t8/100)**2)-t8/100))+tp;			#temperature in deg.C

#Result
print 'the temperature of the bath is %3.2f deg.C'%(t9)

the temperature of the bath is 385.50 deg.C