Question 1. The reading tA and tB of two centigrade scale thermometers A and B graded at ice point 0°C and steam point 100°C are related by the equation tA=l +mtB+n(tB)², where l, m,n are constants. When both are immersed in oil then thermometer A reads 51°C and thermometer B reads 50 °C. determine the reading on thermometer A when thermometer B reads 25°C? Comment on the question "which thermometer is correct".

Thermodynamics Question solution,
Chapter1 Zeroth law of thermodynamics.

Question 1. The reading tA and tB of two centigrade scale thermometers A and B graded at ice point 0°C and steam point 100°C are related by the equation
tA=l +mtB+n(tB)², where l, m,n are constants. When both are immersed in oil then thermometer A reads 51°C and thermometer B reads 50 °C. determine the reading on thermometer A when thermometer B reads 25°C? Comment on the question "which thermometer is correct".

Ans-
tA=l+mtB+n(tB)²
According to question
tA=tB=0°C

tA=l+mtB+n(tB)²
0=l+0+0
l=0
Second step
tA=tB=100°C

tA=l+mtB+n(tB)²
100=0+m×100+n(100)²
m+100n=1             ----------------------------(1)
For
tA=51°C
tB=50°C
tA=l+mtB+n(tB)²
51=0+M50+n(50)² -----------------------------(2)
Now using equation (1) and (2) find m&n
50m+2500n=51
50m+5000n=50
-        -             -
________________
0      -2500n=1
                 n=-1/2500
Put value of 'n' in equation (1)
m+100×(-1/2500)=1
m=26/25

At tB=25°C, l=0

tA=mtB+n(tB)²
     =26/25×25-1/2500×25×25
      =26-1/4
tA=25.75°C
tB=25°C