Given up energy from storaged cheese

[EliFlores]

20,000 Kgs. of fresh cheese (50% water) at 45 °C is placed in a 6°C storage room. We need to calculate the Enthalpy in KW and time for the cheese to reach the final temperature.
I think the cheese mass is 10,000 Kgs. and the water mass is 10,000 Kgs. (Due the 50% of moisture)and Delta T= -39 °C
I have assumed a specific heat for the cheese using this formula:

Q= Mch*Cch(Tf-Ti) + Mw*Cw(Tf-Ti) = 0
so: Cch= Mw*Cw(Tf-Ti)/Mch(Tf-Ti)
= 10,000*4186*(-39)/10,000*(-39)
= 4186 Joules/Kg °C (the same as the water!!!)

Is Correct that calculation?
I have read that a normal specific heat average for cheese is about 2,000 Joules/Kg °C.

 

[IRstuff]

Your answer was a foregone conclusion when you set the Joule Heat to 0.

However, your algebra error hid the fact that you would have gotten a negative specific heat. The Joule heat is what you're trying to calculate and you cannot use it to find the specific heat of the cheese, since you know neither quantity.

You either have to use the value of specific heat that you have or you need to get more data.
TTFN