Thermodynamics: Units in calculations of heat (Q) When calculating an unknown variable in the quantity of heat equation for a system, let's say mixing water of 80 degrees C and ice of -20 degrees C that achieves thermal equilibrium at 10 degrees C (get rid of Q and set the equation equal to zero), you use the equation for quantity of heat (listed below). As you cans see, specific heat (c), is in units of j per kg Kelvin. If you have a delta T (T final - T initial) given in degrees C (like above), do you have to convert your delta T from degrees C to Kelvin so that the this matches the Kelvin in specific heat units? Or does it not matter because there is a temperature of the same magnitude no matter how you look at it? That is, if the delta T in degrees C is 30 degrees C then converting this to Kelvin will be a 30 Kelvin difference also.

 A: In general, units always do matter. For example, if the change in temperature would be given using Farenheit degrees, while the specific heat in J/kg*K, you would have to insert a correction factor.
But fortunately, the Kelvin and Celsius scales are defined so that they are only offset relative to each other, not rescaled. That is, they have different zero points, but one 'tick' has the same length in both systems. Therefore their difference is only apparent when one is interested in absolute temperature, but vanishes, if you are using only temperature differences.
A: Assuming that the final temperature lies between 0 C (273 K) and 100 C (373 K), the heat balance equation in terms of temperatures in centigrade degrees is:  
$$M_{ice}C_{ice}(T_{melting\ C}-T_{ice\ init\ C})+M_{ice}\lambda_{melting}+M_{ice}C_{liquid}(T_C-T_{melting\ C})=M_{liquid}C_{liquid}(T_{liquid\ init\ C}-T_C)$$
where 
$M_{ice}$ = mass of ice initially
$M_{liquid}$ = mass of liquid water initially
$C_{ice}$ = heat capacity of ice
$C_{liquid}$ = heat capacity of liquid water
$\lambda_{melting}$ = heat of melting ice to form liquid water
$T_{ice\ init C}$ = initial temperature of ice
$T_{liquid\ init\ C}$ = initial temperature of liquid water
$T_{melting\ C}$ = melting temperature of ice
$T_C$ = final temperature at equilibrium
Now, in terms of temperatures in degrees K, we have
$T_{ice\ init\ C}=T_{ice\ init\ K}-273$
$T_{liquid\ init\ C}=T_{liquid\ init\ C}-273$
$T_{melting\ C}=T_{melting\ K}-273$
$T_C=T_K-273$
If we substitute these last four relationships into our heat balance equation, we obtain:
$$M_{ice}C_{ice}(T_{melting\ K}-T_{ice\ init\ K})+M_{ice}\lambda_{melting}+M_{ice}C_{liquid}(T_K-T_{melting\ K})=M_{liquid}C_{liquid}(T_{liquid\ init\ K}-T_K)$$
The equation is exactly the same in terms of the Kelvin temperatures as in terms of the Celsius temperatures.  Therefore, it doesn't matter whether you use degrees C or degrees K.
