Melting snow for drinking water: sensible heat, then latent heat
SNC2D Grade 10 Science · Climate Change
On a February trip in Algonquin Park, a camper making drinking water sets a single-burner stove on a packed patch of snow outside the tent. The pot is filled with 1.50 kg of fresh snow, and a thermometer pushed into the middle of it reads −12.0 °C — the same as the afternoon air. The stove, run flat out, delivers 1,200 W of heat into the pot, and the camper watches the thermometer climb steadily at first, then sit unmoving at 0 °C while the snow slumps, turns granular, and slowly disappears into water. Taking the specific heat capacity of ice as 2,100 J/(kg·K) and the latent heat of fusion as 334 kJ/kg, find the heat needed to bring the snow up to 0 °C and how long that takes, then the heat needed to melt it and how long the melt takes.
Every number in this problem is editable, the material included: change any value below and the whole chain recalculates.
- m = 1.5 kg (Snow scooped into the pot)
- T₀ = -12 °C (Snow temperature at the start)
- P = 1,200 W (Heat the stove delivers into the pot)
- c_ice = 2,100 J/(kg·K) (Specific heat capacity of ice)
- L_f = 334 kJ/kg (Latent heat of fusion of ice)
- (a)the heat needed to bring the snow up to 0 °C
- (b)how long the stove takes to deliver it
- (c)the heat needed to melt the snow
- (d)how long the melting takes
Nothing can melt until the snow reaches 0 °C, so the temperature climb is the first heat demand to satisfy. ΔT is a difference, so −12.0 °C to 0 °C is 12.0 K.
Carried onward at full precision, not this rounded figure.
The stove delivers joules at a steady rate, so that heat divided by 1,200 W is the time on the clock.
Carried onward at full precision, not this rounded figure.
Now the phase change. There is no ΔT in this formula at all — the thermometer sits at 0 °C the entire time the ice is disappearing.
Carried onward at full precision, not this rounded figure.
Same stove, same arithmetic as step 2 — and an answer more than thirteen times longer. That gap is the lesson.
Carried onward at full precision, not this rounded figure.
Therefore warming the snow to 0 °C takes 37.8 kJ and 31.5 s of stove time, while melting it demands 501 kJ and another 417.5 s — nearly seven minutes during which the thermometer refuses to move.
Why this order
The chain is split into two heat calculations because they are two genuinely different physical processes, and the classic mistake is to run them together with one Q = mcΔT from −12 °C to some final temperature. That is wrong twice over: it ignores the 501 kJ the melting itself demands, and it uses ice's specific heat for water that no longer exists. Steps 1 and 3 have to be separate lines on the page, and the times in steps 2 and 4 show why anyone should care — 31.5 s to warm the snow, 417.5 s to melt it. Almost seven minutes of a stove's fuel goes into a phase change that the thermometer refuses to acknowledge.
This is also the arithmetic behind spring breakup and behind why Canadian lakes lag the calendar. A lake that has finally reached 0 °C still has to absorb 334 kJ for every kilogram of ice before open water appears, which is why ice can persist through weeks of above-zero air. The same latent heat runs in reverse in the fall: freezing releases it, which is why a large lake holds the first frost off its own shoreline. Joseph Black measured exactly this in Glasgow in the 1760s, and it remains the single largest term in any honest energy budget for a melting ice sheet.
Carried values move at full precision, not the rounded figure shown: chaining rounded numbers compounds error.