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Ice cube in hot water: the melting bill and the energy audit

SPH3U Grade 11 Physics · Energy and Society

For a calorimetry lab, a student lifts a 48.0 g ice cube from an ice-water slurry, blots it dry, and drops it into an insulated foam cup holding 200 g of hot water drawn from an urn at 62.0 °C. Stirring gently, they watch the last sliver of ice disappear and the thermometer settle at 34.0 °C. Before pouring, the student had also wondered how little urn water could melt the cube at all. Take water's specific heat as 4186 J/(kg·K) and ice's latent heat of fusion as 334 kJ/kg.

Given
  • 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 — Urn water temperature
  • T_f = 34 °C — Final temperature, measured
  • L_f = 334 kJ/kg — Latent heat of fusion of ice
  • c_w = 4186 J/(kg·K) — Specific heat of water
Determine
  1. (a)the heat needed to melt the 48.0 g cube
  2. (b)the least mass of 62.0 °C water that could just melt it, the mix ending at 0 °C
  3. (c)the heat the 200 g pour actually gave up cooling to 34.0 °C
  4. (d)the heat that went into warming the melt from 0 °C to 34.0 °C
Step 1 of 4(a) · solve for Heat absorbed or released

The cube comes out of a slurry, so it is already at 0.0 °C and the whole first bill 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.

Rearranged for Q
Q=mLQ = m L
Your values, in your units
Q=(48 g) (334 kJ/kg)Q = \left( 48\ \text{g} \right) \, \left( 334\ \text{kJ/kg} \right)
Converted to base units
Q=(0.048 kg) (334,000 J/kg)Q = \left( 0.048\ \text{kg} \right) \, \left( 334{,}000\ \text{J/kg} \right)
Answer
Q=16.032 kJQ = 16.032\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Latent Heat solver →

Step 2 of 4(b) · solve for Mass

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.

Rearranged for m
m=Qc ΔTm = \frac{Q}{c \, \Delta T}
16.032 kJcarried from step 1
Your values, in your units
m=(16,032 J)(4,186 J/(kg⋅K)) (62 C∘)m = \frac{\left( 16{,}032\ \text{J} \right)}{\left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 62\ \text{C}^{\circ} \right)}
Answer
m=61.773 gm = 61.773\ \text{g}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Step 3 of 4(c) · solve for Heat energy

Now the real pour's supply side. 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.

Rearranged for Q
Q=mcΔTQ = m c \Delta T
Your values, in your units
Q=(200 g) (4,186 J/(kg⋅K)) (28 C∘)Q = \left( 200\ \text{g} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 28\ \text{C}^{\circ} \right)
Converted to base units
Q=(0.2 kg) (4,186 J/(kg⋅K)) (28 C∘)Q = \left( 0.2\ \text{kg} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 28\ \text{C}^{\circ} \right)
Answer
Q=23.442 kJQ = 23.442\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Step 4 of 4(d) · solve for Heat energy

The melt's own bill: 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.

Rearranged for Q
Q=mcΔTQ = m c \Delta T
Your values, in your units
Q=(48 g) (4,186 J/(kg⋅K)) (34 C∘)Q = \left( 48\ \text{g} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 34\ \text{C}^{\circ} \right)
Converted to base units
Q=(0.048 kg) (4,186 J/(kg⋅K)) (34 C∘)Q = \left( 0.048\ \text{kg} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 34\ \text{C}^{\circ} \right)
Answer
Q=6.8316 kJQ = 6.8316\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Therefore melting the cube costs 16.0 kJ, a bare 61.8 g of urn water could just pay it, and the real pour's ledger reads 23.44 kJ given up against 16.03 + 6.83 = 22.86 kJ accounted for — the missing 0.58 kJ, about 2.5%, leaked through the foam and warmed the cup itself.

Why this order

The order is the order of the ledger. The melting bill in part (a) is fixed before any temperature is chosen, so it comes first; part (b) spends it against the steepest possible ΔT to find the least water that could pay it; and parts (c) and (d) then audit the pour that actually happened — supply on one page, the melt's warming on the other, with the melting bill carried over from (a). 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 prices the phase change at zero and it uses a specific heat for a substance that is busy ceasing to exist.

The audit's refusal to balance 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.