you plan on cooking a 20 pound (9.07 kg) turkey for thanksgiving this year. Give

Question

you plan on cooking a 20 pound (9.07 kg) turkey for thanksgiving this year. Given that you are a highly intelligent chemistry student, you started to think aout the thermodynamic properties of your turkey. Specifically, you are curious as to what the change in enthalpy would be for the process of cooking your turkey.

1 Thawed Raw Turkey (s) --> 1 Perfectly Roasted Turkey (S)

What is delta H... delta H= ????/ kJ/kg

You already know that you will defrost your turkey in the refridgerator giving the raw turkey a temperature of 35 degrees F (1.67 degrees C) when the thermal equilibrium with your refridgerator. You are also well aware of the face that the turkey must be cooled to a final temperature of 165 degrees F (73.9 degrees C) to avoid getting the whole famly sick.

While thinking about this you seem to vaguely remember your lab TA mentioning a formula that you could use to solce this problem during the lab wiht the styrofoam cup (since you knew that it was not on the exam, you really did not pay attention). Even with a foggy head after an impossible exam this weekm you somehow manage to recall q=ms(delta T) where s is the specific heat, not the numerous other uses for "s" in your chemistry book. Why cant they just use different letter fir everything (at lease it is not as bad as another "k")!

Explanation / Answer

Using, q = m s dT        -- equation 1   

Where, q= heat change                                                                  , m= mass in kilogram,

s= specific heat (in terms of kJ/kg0C)                       

dT = Final – Initial temperature

Given,

            Mass of turkey = 9.07 kg

            dT = (73.9 – 1.67)0C = 72.230C

            s = 4.12 kJ/kg0C                ; since water content of turkey meat is around 70% (like other animals’ dry mass), it is assumed that the specific heat of turkey is equal to that of water (specific heat of turkey is not provided).

Putting the values in equation 1

                q = 9.07 kg x (4.12 kJ/kg0C) x 72.230C

            or, q = 2699.119532 kJ = 2699 kJ (around)

The amount of heat absorbed (dH) by turkey = 2699 kJ

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