Question

Perform the following conversions:

Answer 1

Answer:

a) $20^0C$ and $293K$

b) $437K$ and $327.2^0F$

c) $-273^0C$ and $459.44^0F$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ , $^0F$and $K$  

These units of temperature are inter convertible.

$t^0C=(t+273)K$

$t^oC=\frac{5}{9}\times (t^oF-32)$

$K-273=\frac{5}{9}\times (^oF-32)$

a)  68°F (a pleasant spring day) to °C and K.

Converting this unit of temperature into $^0C$ and $K$ by using conversion factor:

$t^oC=\frac{5}{9}\times (68^oF-32)$

$t=20^0C$

$20^0C=(20+273)K=293K$

b) 164°C (the boiling point of methane, the main component of natural gas) to K and °F

Conversion from degree Celsius to Kelvins  and Fahrenheit

$164C=\frac{5}{9}\times (t^oF-32)$

$t^0F=327.2$

$164°C=(164+273)K=437 K$

c) 0K (absolute zero, theoretically the coldest possible temperature) to °C and °F.

$K-273=\frac{5}{9}\times (t^oF-32)$

$0-273=\frac{5}{9}\times (t^oF-32)$

$t=-459.4^0F$

$t^K=(0-273)^0C=-273^0C$