A 6-pack of canned drinks is to be cooled from 23°C to 3°C. The mass of each canned drink is 0.355 kg. The drinks can be treated as water, and the energy stored in the aluminum can itself is negligible. The amount of heat transfer from the six canned drinks is kJ.

178 kJ

Explanation:

Assuming no heat transfer out of the cooling device, and if we can neglect the energy stored in the aluminum can, the energy transferred by the canned drinks, would be equal to the change in the internal energy of the canned drinks, as follows:

ΔU = -Q = -c*m*ΔT (1)

where c= specific heat of water = 4180 J/kg*ºC

           m= total mass = 6*0.355 Kg = 2.13 kg

           ΔT = difference between final and initial temperatures = 20ºC

Replacing by these values in (1), we can solve for Q as follows:

Q = 4180 J/kg*ºC * 2.13 kg * -20 ºC = -178 kJ

So, the amount of heat transfer from the six canned drinks is 178 kJ.

-178.92 kJ

The amount or quantity (Q) of heat transferred during a chemical process such as cooling is the product of the mass (m) of the substance involved, the specific heat capacity (c) of the substance and the change in temperature (ΔT) of the substance. i.e

Q = m x c x Δ T   (i)

From the question;

m = total mass of the canned drinks = 6 x 0.355kg = 2.13kg

ΔT = final temperature - initial temperature = 3°C - 23°C = -20°C

Known constant;

c = specific heat capacity of water = 4200J/kg°C

Substitute all these values into equation (i) as follows;

Q = 2.13 x 4200 x (-20)

Q = -178920 J

Divide the result by 1000 to convert it to kJ as follows;

Q = (-178920 / 1000) kJ

Q = -178.92 kJ

Therefore, the quantity of heat transferred from the six canned drinks is -178.92 kJ

Note;

The -ve sign shows that the heat transferred is actually the heat lost in cooling the drinks from 23°C to 3°C

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