Elaborating Ron’s answer via a concrete example, we begin by looking at the Celsius scale

$$ T_ extC1 = 25 mathrm^circ C ,~ T_ extC2 = 100 mathrm^circ C$$

Therefore, we obtain

$$ Delta T_ extC = T_ extC2 - T_ extC1 = 100 mathrm^circ C - 25 mathrm^circ C = 75 mathrm^circ C $$

In a comparable fashion, utilizing the Kelvin scale

$$ T_ extK1 = 298.15 mathrm K = (273.15 + 25) mathrmK,~ T_ extK2 = 373.15 = (273.15 + 100) mathrmK$$

Therefore, we obtain

$$ Delta T_ extK = T_ extK2 - T_ extK1 = (273.15 + 100) mathrmK - (273.15 + 25) mathrmK = 75 mathrmK $$

As you can check out, the consistent offset of 273.15 in between the Celsius range and also the Kelvin range cancels, and the distinction is precisely the very same. The *units* reprimary, yet the *numerical value* of the difference is the exact same.

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edited Mar 31 "15 at 19:20

user7951

answered May 28 "14 at 9:44

Kjetil SonerudKjetil Sonerud

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The temperature in kelvin is equal to the temperature in degrees Celsius plus 273.

So, $Delta T$ will be the very same whether the temperature was reported in $mathrmK$ or $^circmathrmC$. As an example if $T_mathrminitial = 0 ^circmathrmC (273 mathrmK)$ and $T_mathrmfinal = 100 ^circmathrmC (373 mathrmK)$, $Delta T$ is 100 degrees no issue whether you used $^circmathrmC$ or $mathrmK$ in your computation.

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edited Oct 10 "16 at 7:04

user7951

answered Apr 11 "14 at 21:54

ronron

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If your specific heat has actually units of degrees Celsius in it then usage the temperatures in levels Celsius to discover $Delta T$. Although, the difference would certainly be precisely the very same as if you provided kelvin as 1 level distinction is the very same on both scales. Nonetheless, I would discourage making use of degrees Celsius as the SI unit is kelvin and also is the one that"s normally used.

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edited Oct 10 "16 at 7:08

user7951

answered Oct 9 "16 at 22:36

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$egingroup$ Note that both the kelvin and the degree Celsius are SI devices. $endgroup$

–user7951

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