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One equation to calculate a heat change is $q = c_p \cdot m \cdot\Delta T$. I know that $\Delta T$ can be in either degrees Celsius ($^\circ{}\mathrm{C}$) or kelvin ($\mathrm{K}$). However, one thing that confuses me is how the units cancel if $\Delta T$ is in degrees Celsius, since $c_p$ has units of $\mathrm{J\ g^{-1}\ K^{-1}}$. How does this work?

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  • $\begingroup$ Welcome to Chemistry.SE! While in this case it is easy to grasp what you're asking, please include the definitions of all the quantities you are using in formulas in your future questions since this makes it easier for other people to understand your question and answer it. $\endgroup$ – Philipp Apr 12 '14 at 0:37
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Elaborating Ron’s answer through a concrete example, we start by looking at the Celsius scale

$$ T_\text{C1} = 25\ \mathrm{^\circ C} ,~ T_\text{C2} = 100\ \mathrm{^\circ C}$$

Thus, we obtain

$$ \Delta T_\text{C} = T_\text{C2} - T_\text{C1} = 100\ \mathrm{^\circ C} - 25\ \mathrm{^\circ C} = 75\ \mathrm{^\circ C} $$

In a similar fashion, using the Kelvin scale

$$ T_\text{K1} = 298.15\ \mathrm K = (273.15 + 25)\ \mathrm{K},~ T_\text{K2} = 373.15 = (273.15 + 100)\ \mathrm{K}$$

Thus, we obtain

$$ \Delta T_\text{K} = T_\text{K2} - T_\text{K1} = (273.15 + 100)\ \mathrm{K} - (273.15 + 25)\ \mathrm{K} = 75\ \mathrm{K} $$

As you can see, the constant offset of 273.15 between the Celsius scale and the Kelvin scale cancels, and the difference is exactly the same. The units remain, but the numerical value of the difference is the same.

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    $\begingroup$ Or, to put it another way - when using temperature differences $\Delta T$, the degree Celsius and the kelvin are equivalent units since $1\ ^\circ \text{C}$ and $1\ \text{K}$ represent the same temperature interval. $1\ ^\circ \text{C} \equiv 1\ \text{K}$ $\endgroup$ – Ben Norris May 28 '14 at 10:04
  • $\begingroup$ Good clearification! $\endgroup$ – Kjetil Sonerud May 28 '14 at 10:20
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The temperature in kelvin is equal to the temperature in degrees Celsius plus 273.

So, $\Delta T$ will be the same whether the temperature was reported in $\mathrm{K}$ or $^\circ\mathrm{C}$. As an example if $T_\mathrm{initial} = 0\ ^\circ\mathrm{C}\ (273\ \mathrm{K})$ and $T_\mathrm{final} = 100\ ^\circ\mathrm{C}\ (373\ \mathrm{K})$, $\Delta T$ is 100 degrees no matter whether you used $^\circ\mathrm{C}$ or $\mathrm{K}$ in your computation.

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  • $\begingroup$ Alright now I would want the person who flagged this as VLQ enlighten me of the issue this answer has. This answer is what the OP looked for! $\endgroup$ – M.A.R. Mar 31 '15 at 20:36
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If your specific heat has units of degrees Celsius in it then use the temperatures in degrees Celsius to find $\Delta T$. Although, the difference would be exactly the same as if you used kelvin as 1 degree difference is the same on both scales. Nevertheless, I would discourage using degrees Celsius as the SI unit is kelvin and is the one that's normally used.

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  • $\begingroup$ Note that both the kelvin and the degree Celsius are SI units. $\endgroup$ – Loong Oct 10 '16 at 7:15

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