My Thermal Mug Math Model

My approach for this problem was to start simple and work my way up to what I am actually solving and working to get.  In this case, making my math model for the Thermal Mug project, I started with some scientific concepts and laws and then worked on deriving an algebraic equation for the equilibrium temperature in the mug after the warm liquid was put inside the cold mug.  To do this I set up my MATLAB Math Model with functions of relevant variables such as initial temperature, heat capacity of water, the mug material, mass values, etc.


My Math Model:

My math model models the temperature change of a chilled mug cooling a room temperature liquid or beverage. 

I used results found online for the specific heat capacities of the glass and the water as well as normal temperatures for rooms and freezers:
Glass: 0.84 [KJ/(kg*K)] = cg
Water: 4.18 [KJ/(kg*K)] = cl
Temp of frozen mug: -13 C
Temp of room temp water: 25 C

Assumptions:
-No work on system
-No loss of temperature to air

State one: Cold mug and warm liquid
State two: Equilibrium state

Goal: Find the temperature at state two

Equations:
deltaE=mc deltaT
deltaE=0 so deltaEmug=deltaEwater

deltaEmug= (mglass)(cglass) (T2-Tmug)
deltaEwater= (mglass)(cglass) (Twater-T2)

so: (mglass)(cglass) (T2-Tmug) = (mglass)(cglass) (Twater-T2)
Solve for T2 to get T2= ((mwater)(cwater)(Twater)+(mglass)(cglass)(Tmug))/((mglass)(cglass)+(mwater)(cwater))

Then checking units reveals the answer in Kelvins which works.



My MATLAB model:

thermalmug.m

% thermalmug.m calculates the final temperature of a liquid in a chilled

% glass. This is only correct for an an idealized case where there is no

% loss of temperature to the surrounding air.

 

clear

cg= .84; % Specific heat capacity of glass [KJ/(kg*K)]

cl= 4.18; % Specific heat capacity of water [KJ/(kg*K)]

 

Tmug= -13; % Temperature of mug [Celcius]

Tliq= 25; % Temperature of water [Celcius]

mg= .5 % Mass of a glass mug

ml= .25; % Mass of water

Tm= Tmug + 273.15; % Mug temp in [kelvin]

Tl= Tliq + 273.15; % Water temp in [kelvin]

 

% Temperature of water at equilibrium in [kelvin]

T2 = (ml*cl*Tl + mg*cg*Tm)/(mg*cg + ml*cl);

TempC = T2 - 273.15 %Final Celcius

TempF = TempC*1.8+32 %Final Fahrenheit

The results:

TempC = 14.1058 C

TempF = 57.3904 F

 

As you can see this would be a reasonable drinking temperature, not a cold as I would like, but it makes sense that just freezing a glass mug and adding a warm liquid, the temperature would still be closer to that of the intitial liquid than being closer to freezing.