Respuesta :
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.
-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: The boiling point of water is [tex]400.34^oA[/tex]
Explanation:
To convert the units of temperature, we use the equation:
[tex]\frac{\text{X-LFP}}{\text{UFP-LFP}}=\frac{\text{Y-LFP}}{\text{UFP-LFP}}[/tex]
where,
X = temperature in [tex]^oA[/tex]
Y = temperature in [tex]^oC[/tex]
LFP = lower fixed point
UFP = upper fixed point
LFP of [tex]^oA[/tex] is 0°A
UFP of [tex]^oA[/tex] is 100°A
LFP of [tex]^oC[/tex] is -77.75°A
UFP of [tex]^oC[/tex] is -33.35°A
Normal boiling point of water = [tex]100^oC[/tex]
Putting values in above equation, we get:
[tex]\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[/tex]
Hence, the boiling point of water is [tex]400.34^oA[/tex]
