Question

You wish to chill 12 cans of soda at 25.0 ∘C down to 5.0 ∘C before serving them to guests. Each can has a mass of 0.354 kg, and the specific heat of soda is the same as that of water.

A.)How much thermal energy must be removed in order to chill all 12 cans?

B.)If the chilling process requires 4.4 h, what is the average rate (in W) at which thermal energy is removed from the soda?

Answer 1

Number of cans = n = 12

Mass of each can = m = 0.354 kg

Initial temperature of the soda = T₁ = 25 ^oC

Final temperature of the soda = T₂ = 5 ^oC

Specific heat of the soda = C = 4186 J/(kg.^oC) (Same as water)

Thermal energy to be removed = Q

Q = nmC(T₁ - T₂)

Q = (12)(0.354)(4186)(25 - 5)

Q = 355642.56 J

Time taken for the chilling process = T = 4.4 hours = 15840 sec

Average rate at which thermal energy is removed from soda = P

Q = PT

355642.56 = (15840)P

P = 22.45 W

A) Thermal energy to be removed in order to chill all 12 cans = 355642.56 J

B) Average rate at which thermal energy is removed from the soda for the chilling process to take place in 4.4 h = 22.45 W