In a thermometer the ice point is marked as 10 degree Celsius,and steam point as 130 degree Celsius .What will be the reading of this scale when it is actually 40 degree Celsius?

1 Answer
Mar 18, 2018

Relationship between two thermometer is given as,

(C- 0)/(100-0)=(x-z)/(y-z)

where, z is the ice point in the new scale and y is the steam point in it.

Given, z=10^@C and y=130^@C

so, for C=40^@C,

40/100=(x-10)/(130-10)

or,x=58^@C