Question

You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point of ammonia (-33.35∘c) is 100 ∘a. part a what would be the boiling point of water in ∘a?

Answer 1

First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

Answer 2

Answer: The boiling point of water is
`400.34^oA`

Step-by-step explanation:

To convert the units of temperature, we use the equation:

`\frac{\text{X-LFP}}{\text{UFP-LFP}}=\frac{\text{Y-LFP}}{\text{UFP-LFP}}`

where,

X = temperature in
`^oA`

Y = temperature in
`^oC`

LFP = lower fixed point

UFP = upper fixed point

LFP of
`^oA` is 0°A

UFP of
`^oA` is 100°A

LFP of
`^oC` is -77.75°A

UFP of
`^oC` is -33.35°A

Normal boiling point of water =
`100^oC`

Putting values in above equation, we get:

`(T(^oA)-0^oA)/(100^oA-0^oA)=(100^oC-(-77.75^oC))/(-33.35^oC-(-77.75^oC))\\\\(T(^oA))/(100^oA)=(177.75^oC)/(44.4^oC)\\\\T(^oA)=400.34^oA`

Hence, the boiling point of water is
`400.34^oA`