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The temperature scale of a thermometer is given by : t = a ln p + b. If at the ice point and steam point the thermometric properties are found to be 2 and 8 respectively, what will be the temperature corresponding to the thermometric property of 4 on Celsius scale ?
t = a ln p + b
At ice point, 0 = a ln 2 + b
At steam point, 100 = a ln 8 + b
Which gives, a=\frac{100}{\ln 4}=72.135
b = – 50
t = 72.135 ln p – 50
At p = 4, t = 72.135 ln 4 – 50 = 50°C