Problem 15 IYPT 2027

15. Cold drink

Thermodynamics

Problem Description & Analysis

15. Cold Drink

A can or bottle of drink at room temperature can be cooled in a cooler containing ice. Adding salt and alcohol to the ice can increase the rate of cooling and affect the drink’s final temperature. Investigate how the relevant parameters affect the lowest achievable temperature and the maximum cooling rate.

What happens when you add ice to a drink — or take a cold can out of a fridge — and what controls how quickly it warms up? Three heat-transfer mechanisms compete, and a phase change pins the temperature while ice remains.

Newton’s Law of Cooling

For a beverage of heat capacity C_p and overall heat transfer coefficient UA:

C_p · dT/dt = −UA · (T − T_amb)

Solution: T(t) = T_amb + (T₀ − T_amb) · e^{−t/τ}, τ = C_p/(UA).

Two-Stage Warming (Ice Present)

Stage 1 (ice present): T = 0°C (constant) for 0 < t < t_melt
Stage 2 (no ice): T(t) = T_amb − ΔT₀ · e^{−(t−t_melt)/τ}

where t_melt = m_ice · L_f / (UA · T_amb), L_f = 334 kJ/kg.

Heat Transfer Mechanisms

Conduction: Q̇_cond = k · A / d · ΔT — dominant for thin aluminium can
Convection: Q̇_conv = h_c · A · ΔT (h_c ≈ 5–10 W/m²·K still air, ≈50 W/m²·K moving air)
Radiation: Q̇_rad ≈ 4εσT³ · A · ΔT — small correction at ambient T
Condensation: Q̇_latent = h_fg · dm/dt — can be 20–40% of total

Condensation (Latent Heat)

Moist air at T_amb cools below the dew point T_d near the cold surface. Condensation rate:

dm/dt = h_c/(c_p·Le^{2/3}) · (w_∞ − w_s(T_wall))

where Le = α/D_AB is the Lewis number, w is specific humidity, and h_fg = 2.5 MJ/kg.

References

Incropera et al. (2007). Fundamentals of Heat and Mass Transfer, 6th ed. — Ch. 2, 7
Cengel & Ghajar (2020). Heat and Mass Transfer, 5th ed.