Question

I have been studying and there is a question in my book asking:

A faulty thermometer reads freezing and boiling point of a liquid as -5°C and 95°C respectively on the Celsius scale. What is the correct value of temperature when it reads 60°C on the faulty thermometer?

(a) 60°C (b) 65°C (c) 64°C (d) 62°C

The solution on back of the textbook is as follows:

(b) 65°C

For all temperature scales,

(X - LFP)/(UFP - LFP) = constant

where LFP is Lower fixed point, and UFP is Upper fixed point

[X - (-5)]/[95 - (-5)] = (C - 0)/(100 - 0)

=> (60 + 5)/(95 + 5) = C/100

=> C = 65°C

In this, I do not understand from where the C - 0 and 100 - 0 came in step 2

EDIT: Fixed typo from -95°C to 95°C

Answer 1

The equation in the solution corrects the temperature read by a faulty thermometer. By creating a proportion between the actual Celsius scale and the faulty readings, we determine the true temperature. The true reading when the faulty thermometer shows 60°C is 65°C.

Step-by-step explanation:

The student is confused about the equation (X - LFP)/(UFP - LFP) = (C - 0)/(100 - 0) used in the textbook's solution to correct the temperature read by a faulty thermometer. In a correct thermometer, the Lower Fixed Point (LFP) for the Celsius scale is 0°C (freezing point of water) and the Upper Fixed Point (UFP) is 100°C (boiling point of water). The 'C - 0' in the equation comes from the concept that we are looking for the real temperature value 'C' that the faulty thermometer would have shown if it was accurate. Since the LFP is actually at -5°C and UFP is at 95°C for the faulty thermometer, by setting up a proportion between the real scale and the faulty scale, we can solve for 'C', which turns out to be 65°C.