First slide

Temperature and heat-thermal properties of matter

Question

The relation between centigrade scale & Fahrenheit scale is

Easy

A

$\frac{C}{100} = \frac{K - 273}{100}$
B

$F = 9 C$
C

$\frac{C}{5} = \frac{F - 32}{9}$
D

$C = \frac{F}{9}$

Solution

$r e l a t i o n b e t w e e n a n y t w o s c a l e s i s a c c o r d i n g t o \frac{s c a l e - l o w e r f i x e d p o i n t}{u p p e r f i x e d p o i n t - l o w e r f i x e d p o i n t} h e n c e f o r c e n t i g r a d e a n d F a h r e n h e i t s c a l e \frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32} \Rightarrow \frac{C}{5} = \frac{F - 32}{9}$

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