The readings ta and tb of two celsius thermometers a and b agree at the ice point and steam point, but else where they are related by the equation ta = l+mtb + ntb 2 where l, m and n are constants, when both the thermometers are immersed in a well stirred bath, a registers 51oc whereas b registers 50oc. Evaluate the reading on b when a registers 25oc.
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Explanation:
At freezing point , both ta =tb=0, so
0=l+0+0
l=0
At boiling point both ta=tb=100° C , so
100=0+(100)m+(100)2 n
1=m+100n
m =1-100n
Now at ta=51 °Cand tb=50°C
51=0+50m+(50)2 n
now put value of m in above equation
51=0+50(1-100n)+2500n
51=50-5000n +2500n
1= -2500n
n= -1/2500
put value of n in m
m=1-100/2500
m=26/25
now at ta=25
25=0+{(26/25)tb}+{(-1/2500)tb2}
now solve it and get answer
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