Physics, asked by HamsaThalvar9168, 1 year ago

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.

Answers

Answered by shrishti2984
8

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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