I remove 250 grams of ice cubes from a freezer and drop it into a bottle of soda

Question

I remove 250 grams of ice cubes from a freezer and drop it into a bottle of soda. The soda is at room temperature, 20 degree C. Assume this is a new 1 L bottle. The freezer is set to a chilly7 degree F. Neglect any heat exchange in moving the ice to the soda, so the initial temperature of the cube is that of the freezer. Further neglect any exchange between the system and its environment. Assume the thermal properties of soda are the same as water. Find the Equilibrium condition. Report mass and temperature for each phase that exists once equilibrium has been achieved.

Explanation / Answer

Equilibrium condition is

heat gained by ice = heat lost by soda

mi*Si*dTi = mW*Sw*dTw

mi is the mass of the ice = 250 g = 0.25kg

Si is the specific heat of ice = 2030 J/kg*C

dTi is the change in temparature of ice

mw is the mass of the water or soda = density*volume = (1000 kg/m^3)*(10^-3 m^3) = 1 kg.........

since 1L = 10^-3 m^3

Sw is the specifice heast capacity of water = 4186 J/kg*C

dTw is the change in temparature of soda

Converting 7F into celsius

then C/5 =(F-32)/9 = (7-32)/9 = -2.77

C = -13.88 degree C

then heat gained by ice at -13.88 degree C is Q1 = (mi*dT1*Si) +(mi*Lf) + (mi*Sw*T) = (0.25*13.88*2030) +(0.25*333550)+ (0.25*4186*T) = 90431.6+1046.5*T

T is the equlibrium temparature

and heat lost by the water = mw*Sw*dT2 = 1*4186*(20-T)

then 1*4186*(20-T) = 90431.6+1046.5*T

83720-90431.6 = (1046.5+4186)*T

T = -1.28 degree C

--------------------------------------------------------

different phases of ice

phase 1 : only ice

mass of ice is mi = 0.25 kg

temparature change is dT = 13.88 degree C

Phase 2 : ice to water

mass is m = 0.25 g

temparature change is zero

phase 3:

only water

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