Keeping Vaccines Cold With 1.5 kg of Ice
A mobile vaccination team will run a clinic in a sports hall from 09:00 to 18:00. Their insulated cool box is packed at 08:30 with 1.5 kg of ice taken from a freezer at −5.0 °C. The box is well sealed but not perfect: a test with a data logger shows it gains heat at an average rate of 14 W from the room, which is at 24 °C. The vaccines must be kept between 2 °C and 8 °C. The team expects the lid to be opened about 20 times during the day to take out and replace trays; a rough estimate from the logger trace is that each opening admits about 15 kJ of heat. Data: specific heat capacity of ice = 2.1 × 10³ J kg⁻¹ K⁻¹, specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹, specific heat capacity of water = 4.18 × 10³ J kg⁻¹ K⁻¹.
1. Calculate the total energy the ice can absorb before it has all melted, and hence the time the box would hold near 0 °C if it were never opened.
Read the worked answer
Two stages, and both must be counted:
Warming the ice from −5.0 °C to 0 °C: Q₁ = mcΔθ = 1.5 × 2100 × 5.0 = 15 750 J
Melting it at 0 °C: Q₂ = mL = 1.5 × 3.34 × 10⁵ = 501 000 J
Total Q = 15 750 + 501 000 = 5.17 × 10⁵ J
Time with a steady 14 W leak: t = Q/P = 516 750/14 = 36 900 s ≈ 10.3 h
So on paper the ice survives past 18:00 with about 1.3 h to spare. Note how lopsided the two terms are: the latent heat is 32 times the sensible heat from warming the ice, which is why the initial −5 °C gives only a few percent of extra cooling time and why pre-freezing to −18 °C would still not transform the performance.
2. Including the 20 lid openings, will the box last until 18:00? Quantify, and state the smallest change that would make the plan safe.
Read the worked answer
Heat admitted by openings: 20 × 15 000 = 300 000 J.
This heat comes out of the same ice budget, so the energy left to absorb the steady leak is: 516 750 − 300 000 = 216 750 J
t = 216 750/14 = 15 480 s = 4.3 h
The ice is fully melted by about 12:50 - roughly five hours short. The plan is not safe. Once the ice is gone the box has essentially no thermal buffer and the contents drift towards room temperature.
Minimum ice needed: energy required = leak over 9 h + openings = (14 × 32 400) + 300 000 = 453 600 + 300 000 = 753 600 J Energy per kg of ice = (2100 × 5) + 3.34 × 10⁵ = 344 500 J kg⁻¹ m = 753 600/344 500 = 2.19 kg
So pack about 2.5 kg to leave a margin, and/or cut the number of openings, and/or pre-chill the box so the leak rate falls. Both the mass and the openings matter: halving the openings saves more time than doubling the ice costs money.
3. A colleague proposes replacing the ice with 1.5 kg of water already at 0 °C, arguing that "it is the same cold and it will not make the box wet". Evaluate this, and say whether the meltwater should be drained during the day.
Read the worked answer
Water at 0 °C can only absorb heat by warming up. To stay within the 8 °C limit it can rise 8 K: Q = mcΔθ = 1.5 × 4180 × 8 = 50 000 J, giving 50 000/14 = 3 600 s ≈ 1 h of protection - a factor of ten worse than the ice.
The physics is that melting is an isothermal process: while ice is melting, the mixture sits at 0 °C and absorbs a large amount of energy (3.34 × 10⁵ J per kg) without its temperature changing. That plateau is exactly what a cold chain needs, because it means the box temperature is pinned near 0 °C no matter how variable the heat load is. Water warming gives a steadily rising temperature instead.
On draining: don't. The meltwater is at 0 °C and still has 1.5 × 4180 × 8 ≈ 50 kJ of usable capacity before it reaches the 8 °C limit, worth about an extra hour. Draining it throws that away and also removes the liquid thermal contact between the remaining ice and the trays. The right fix for wetness is to keep the ice in sealed bags, not to drain the box.