# Celsius to Fahrenheit, Kelvin Example Problems

Relation connecting the scales of temperature:

If C is the temperature of a body on the Celsius scale, F is the temperature on the Fahrenheit scale and Kis the temperature on Kelvin’s scale, then,

C/100 = (F - 32)/180 = (K - 273)/100
Or C/5 = (F - 32)/9 = (K - 273)/5
Intervals of 9° on Fahrenheit scale is equal to intervals of 5° on Celsius scale,
That is, 9 F° = 5 C° (Difference in readings are given in F° or C°)
1F ° = (5/9) C°

Celsius to Fahrenheit, Kelvin Example Problems:

1. Temperature of the human body is 98.4°F. Find the corresponding temperatures on the Celsius scale and Kelvin scale.

Answer: C/5 = (F – 32)/9 ; C/5 = (98.4 – 32)/9 = 66.4/9
C = (5 X 66.4)/9 = 36.9° C
On the Kelvin scale = (273 + t°C) = 273 + 36.9 = 309.9 K

2. At what temperature do Celsius and Fahrenheit scales coincide?

Answer: Let X be the temperature at which Celsius and Fahrenheit scales are equal.
C/5 = (F - 32)/9; X/5 = (X – 32)/9 :
Cross multiplying;
9X = 5(X - 32) = 5X -160
Therefore, X = -40

3. At what temperature do the Kelvin and Fahrenheit scales coincide

Answer: Let X be the temperature at which the Kelvin and Fahrenheit scales coincide.
(K – 273)/5 = (F - 32)/9; (X - 273)/5 = (X – 32)/9;
Cross multiplying, 9X – 2457 = 5X – 160
4X = 2297;
Therefore, X = 2297/4 = 574.25

4. If the change of temperature of a body is 20°C, find the change of temperature of the body in Fahrenheit scale.

Answer: 100C° = 180 F° ; 1C° = (9/5) in F° (C° or F° means a change of :)
Change of 20°C = (9/5) x 20 = a change of 36°F

5. A constant volume gas thermometer using helium records a pressure of 20 kPa at the triple point of water and a pressure of 14.3 kPa at the temperature of ‘dry ice’ (solid CO2). What is the temperature of ‘dry ice’?

Answer: P1 = 20 kPa;  T1 = T1r = 273 K:  P2 = 14.3 kPa, T2 = ?;
According to Charle’s law, at constant volume,
P1/T1 = P2/T2;
T2 = P2/P1 x T1 = (14.3/20) x 273 = 195 K