**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 CO**

_{2}). What is the temperature of ‘dry ice’?

**Answer:**P

_{1}= 20 kPa; T

_{1}= T

_{1r}= 273 K: P

_{2}= 14.3 kPa, T

_{2}= ?;

According to Charle’s law, at constant volume,

P

_{1}/T_{1}= P_{2}/T_{2};
T

_{2}= P_{2}/P_{1}x T_{1}= (14.3/20) x 273 = 195 K
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