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Which Method Is Correct for Converting Kelvin to Celsius?
A)

Question 1

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 )

Correct Answer:

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