Question

A 6 -pack canned drink is to be cooled from \(18^{\circ} \mathrm{C}\) to \(3^{\circ} \mathrm{C} .\) 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 6 canned drinks is \((a) 22 \mathrm{kJ}\) (b) \(32 \mathrm{kJ}\) \((c) 134 \mathrm{kJ}\) \((d) 187 \mathrm{kJ}\) \((e) 223 \mathrm{kJ}\)

Short answer

Question: Determine the amount of heat transfer needed to cool a 6-pack canned drink (treated as water) from 18°C to 3°C. The mass of each canned drink is 0.355 kg and the specific heat capacity of water is 4.18 kJ/kg°C. Ignore the energy stored in the aluminum can itself. a) 50 kJ b) 100 kJ c) 134 kJ d) 200 kJ

Step-by-step solution

Calculate the Change in Temperature

Since we need to cool the drinks from 18°C to 3°C, the change in temperature is: ΔT = Initial temperature - Final temperature = 18°C - 3°C = 15°C

Calculate the Mass of the 6 Canned Drinks

We know the mass of each canned drink is 0.355 kg. To calculate the total mass for the 6-pack, we multiply the mass of each drink by 6: m_total = 6 × 0.355 kg = 2.13 kg

Use the Heat Transfer Formula

Now we can use the heat transfer formula to find the amount of heat transfer for the 6-pack canned drink: q = m × c × ΔT We know: - m_total = 2.13 kg - c (specific heat capacity of water) = 4.18 kJ/kg°C - ΔT = 15°C Calculating q: q = (2.13 kg) × (4.18 kJ/kg°C) × (15°C) = 133.89 kJ Since the amount of heat transfer is close to 134 kJ, the correct answer is: \((c) 134 \mathrm{kJ}\)