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Which Method Is Correct for Converting Kelvin to Celsius?
A) T(C)=(5C9 K)T(K)+32\mathrm{T}\left({ }^{\circ} \mathrm{C}\right)=\left(\frac{5^{\circ} \mathrm{C}}{9 \mathrm{~K}}\right) \mathrm{T}(\mathrm{K})+32

Question 27

Multiple Choice

Which method is correct for converting kelvin to Celsius?


A) T(C) =(5C9 K) T(K) +32\mathrm{T}\left({ }^{\circ} \mathrm{C}\right) =\left(\frac{5^{\circ} \mathrm{C}}{9 \mathrm{~K}}\right) \mathrm{T}(\mathrm{K}) +32
B) T(C) =(9C5 K) T(K) +273.15\mathrm{T}\left({ }^{\circ} \mathrm{C}\right) =\left(\frac{9^{\circ} \mathrm{C}}{5 \mathrm{~K}}\right) \mathrm{T}(\mathrm{K}) +273.15
C) T(C) =5C9 K( T( K) 273.15) \mathrm{T}\left({ }^{\circ} \mathrm{C}\right) =\frac{5^{\circ} \mathrm{C}}{9 \mathrm{~K}}(\mathrm{~T}(\mathrm{~K}) -273.15)
D) T(C) =1C1 K( T( K) 273.15) \mathrm{T}\left({ }^{\circ} \mathrm{C}\right) =\frac{1^{\circ} \mathrm{C}}{1 \mathrm{~K}}(\mathrm{~T}(\mathrm{~K}) -273.15)
E) T(C) =1C1 K( T( K) +273.15) \mathrm{T}\left({ }^{\circ} \mathrm{C}\right) =\frac{1^{\circ} \mathrm{C}}{1 \mathrm{~K}}(\mathrm{~T}(\mathrm{~K}) +273.15)

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