Ice cube in hot water: melting heat and the energy balance
SPH3U Grade 11 Physics · Energy and Society
The calorimetry bench holds a nested pair of foam cups with a lid and thermometer port, a stirring rod, a hot-water reservoir, and a tray of ice cubes sitting in their own ice-water slurry. The student lifts a 48.0 g ice cube from the slurry with tongs and blots it dry on a paper towel, so no clinging meltwater rides along as extra mass. The cube goes into the insulated foam cup, which holds 200 g of hot water just drawn from the reservoir at 62.0 °C, and the lid goes on. Stirring gently, the student watches through the port as the cube shrinks and rounds off, until the last sliver of ice disappears and the falling thermometer settles at 34.0 °C. Before pouring, the student had also wondered how little hot water could melt the cube at all — the limiting case where the mix would end right at the freezing point. Take water's specific heat as 4,186 J/(kg·K) and ice's latent heat of fusion as 334 kJ/kg.
Every number in this problem is editable, the material included — change any value below and the whole chain recalculates.
- m_i = 48 g — Mass of the ice cube (at 0.0 °C)
- m_w = 200 g — Hot water actually poured
- T_h = 62 °C — Hot water temperature
- T_f = 34 °C — Final temperature, measured
- L_f = 334 kJ/kg — Latent heat of fusion of ice
- c_w = 4,186 J/(kg·K) — Specific heat of water
- (a)the heat needed to melt the 48.0 g cube
- (b)the least mass of 62.0 °C water that could just melt it, the mix ending at 0 °C
- (c)the heat the 200 g of hot water actually released cooling to 34.0 °C
- (d)the heat that went into warming the melt from 0 °C to 34.0 °C
The cube comes out of a slurry, so it is already at 0.0 °C and the entire first quantity is the phase change: Q = mL, no ΔT anywhere in it. The blotting matters too — clinging meltwater would ride in as mass that needs no melting and quietly inflate m.
Carried onward at full precision, not this rounded figure.
The limiting case: if the mix ends exactly at 0 °C, every joule the hot water surrenders on its 62.0 K slide goes to melting and none to warming melt. Heat lost equals heat gained is not a law of nature here — it is what the foam cup is FOR, and the answer lands eerily close to the cube's own mass.
Carried onward at full precision, not this rounded figure.
Now the heat the real 200 g actually released. The hot water's ΔT is 62.0 − 34.0 = 28.0 K — the drop it actually suffered — never the 34.0 on the thermometer's face. Reading a final temperature as a temperature CHANGE is the most reliably marked-down slip in calorimetry.
Carried onward at full precision, not this rounded figure.
The meltwater's own heat demand: 48.0 g of brand-new liquid water climbs from 0 °C to 34.0 °C, using water's specific heat now, not ice's — the solid it came from no longer exists. This is the term students forget entirely when they treat "melt the ice" as the whole story.
Carried onward at full precision, not this rounded figure.
Therefore melting the cube requires 16.0 kJ, a minimum of 61.8 g of hot water could supply exactly that, and the energy balance reads 23.44 kJ released against 16.03 + 6.83 = 22.86 kJ absorbed — the 0.58 kJ residual, about 2.5%, leaked through the foam and warmed the cup itself.
Why this order
The order is the order of the energy balance. The heat of melting in part (a) is fixed before any temperature is chosen, so it comes first; part (b) divides it by the steepest possible ΔT to find the least water that could supply it; and parts (c) and (d) then check the mixture that actually happened — heat released on one side, the meltwater's warming on the other, with (a)'s heat of melting carried over. Nothing in the chain ever needs two heats added inside a formula: each term is computed on its own line and the totalling happens where a lab write-up does it, in the closing statement. That separation is deliberate. The classic wreck is one grand Q = mcΔT run from "ice" to 34 °C, which is wrong twice — it counts the phase change as zero heat and it uses a specific heat for a substance that is busy ceasing to exist.
The balance's refusal to close is the second lesson, and it is a feature, not an error. Heat lost equals heat gained is an idealization the foam cup only approximates: 578 J — about 2.5% of the supply — went into the cup wall, the thermometer stem and the air, and a good write-up names that residue instead of hiding it. Real calorimetrists chase exactly this term with vacuum jackets and stirring corrections. Part (b)'s near-coincidence is worth a pause of its own: 61.8 g of 62 °C water to melt a 48.0 g cube says melting ice is nearly as expensive as heating the same water from 0 to 80 °C — the same latent-heat enormity that lets one tray of cubes hold a whole pitcher at the edge of 0 °C all afternoon.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.