# 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. • Did you try both ways (converting to K or not)? What did you get? Also, above formula fails if there is a change in state of matter (ice -> fluid water). – Jasper Dec 6 '18 at 16:34
• Yes it does fail. I believe that is why it's important that the final temp of 20 degrees C is listed. This implies that the system reached thermal equilibrium, so you can essentially get rid of Q and set the equation equal to zero. Yes, you are correct. That is a flaw in my question. I guess I'm just wondering if units of temp measurement need to be converted or if their magnitude is all that matters. – Jonathan L. Dec 6 '18 at 16:41
• This is a homework problem, so I'm using different numbers and being discrete about what I'm solving for. I don't want my problem solved. I just want to know if the units in specific heat and temperature change matter. My hw problem is not the picture above! – Jonathan L. Dec 6 '18 at 16:45
• Then make up some numbers with temperatures in celsius, calculate heat (glossing over the unit mismatch), convert to kelvin (be happy about matching units), calculate again and compare. – Jasper Dec 6 '18 at 17:00
• Forget the units for a moment. Your heat equation only applies to sensible heat. If the ice reaches 0 C and changes phase to water you will need latent heat as well. – Bob D Dec 6 '18 at 19:06

## 2 Answers

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.

• Thank you. Very well done explanation and example. – Jonathan L. Dec 7 '18 at 17:37

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.