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 $400.34^oA$

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:

$\frac{T(^oA)-0^oA}{100^oA-0^oA}=\frac{100^oC-(-77.75^oC)}{-33.35^oC-(-77.75^oC)}\\\\\frac{T(^oA)}{100^oA}=\frac{177.75^oC}{44.4^oC}\\\\T(^oA)=400.34^oA$

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