Choosing a Cold Pack: Heat Capacity vs. Thermal Conductivity
A student athletic trainer is choosing a reusable cold pack for a 20-minute application to a sprained ankle. Two options sit on the bench: Pack A contains 250 g of water at 4 °C; Pack B contains a 250 g aluminum block at 4 °C. Both would be pressed directly against skin, where the tissue surface warms to about 30 °C beneath the pack. Reference values given in the training manual: for water, c = 4.18 J/(g·°C) and k = 0.6 W/(m·K); for aluminum, c = 0.90 J/(g·°C) and k = 205 W/(m·K). The manual also states that 250 g of ice absorbs 334 J per gram while melting at 0 °C. A note in the margin reads: "Aluminum is better because metal gets colder faster."
1. Calculate how much heat each pack absorbs while warming from 4 °C to 30 °C. Which pack holds the larger "reservoir" of cooling, and by what factor?
Read the worked answer
Use q = m·c·ΔT with ΔT = 30 − 4 = 26 °C.
- Water: q = 250 g × 4.18 J/(g·°C) × 26 °C = 27,170 J ≈ 27.2 kJ
- Aluminum: q = 250 g × 0.90 J/(g·°C) × 26 °C = 5,850 J ≈ 5.85 kJ
Ratio: 27,170 ÷ 5,850 ≈ 4.6
The water pack absorbs about 4.6 times more heat over the same temperature rise, so it is the larger reservoir. The margin note confuses rate of heat flow with capacity to absorb heat; the calculation isolates capacity only, and water wins decisively because of its high specific heat.
2. Using the conductivity values, which pack is more likely to injure skin during a 20-minute direct application? Explain the trade-off between conductivity and heat capacity, and state one practical change.
Read the worked answer
Heat flux across the skin–pack interface scales with conductivity: q/A = k·ΔT/L. For the same temperature difference and contact thickness, aluminum (k = 205) moves heat roughly 340 times faster than water (k = 0.6).
So the two properties pull in opposite directions:
- Aluminum: very fast heat extraction, small reservoir. The skin surface chills almost instantly even though the block as a whole barely warms, producing a steep surface-to-deep gradient - the condition that produces cold injury (frostbite-type damage) in minutes.
- Water: slower extraction, large reservoir. It sustains a moderate, tolerable cooling for the full 20 minutes.
Judgment: the aluminum pack carries the higher injury risk despite its smaller total cooling; the limiting factor is the rate at contact, not total joules. Practical change: never apply metal directly to skin - wrap either pack in a cloth barrier and check skin every 5 minutes. (This same flux difference is why metal at 4 °C feels colder than water at 4 °C even at identical temperature.)
3. A third option is 250 g of ice at 0 °C. Compare its total heat absorbed (0 °C ice → 30 °C water) with Pack A, and discuss one safety advantage and one safety drawback of ice.
Read the worked answer
Ice must first melt, then warm:
- Latent heat of fusion: 250 g × 334 J/g = 83,500 J
- Sensible warming 0 → 30 °C: 250 g × 4.18 × 30 = 31,350 J
- Total ≈ 114,850 J ≈ 114.9 kJ
Compare with Pack A (27.2 kJ): the ice pack holds about 4.2 times more cooling, and most of that (83.5 kJ, ~73%) is stored in the phase change rather than in a temperature rise.
- Advantage: while melting, the pack is pinned at 0 °C, so the delivered temperature is predictable and cannot plunge toward the −60 °C range that a low-capacity material could approach; the long plateau also outlasts water.
- Drawback: 0 °C ice still causes frostbite with prolonged direct contact, so the same cloth barrier and skin checks are required, and the 20-minute limit is not extended just because more cooling remains.