Latent heat — phase transitions: Exercises — Physics deep dive

Figure — Latent heat — phase transitions

The figure above is your map: each slope costs $m c Δ T$ , each flat step costs $m L$ . Keep pointing back to it — every problem is just "which pieces of this staircase am I climbing?"

Level 1 — Recognition

Goal: name the right formula and read off the answer with one substitution.

L1.1

Q: $150$ g of ice sits exactly at $0° C$ . How much heat fully melts it?

Recall Solution

The ice is already at the melting point, so no warming term — this is a pure plateau, use $Q = m L_{f}$ . $Q = m L_{f} = 0.150 \times 3.34 \times 1 0^{5} = 5.01 \times 1 0^{4} J = 50.1 kJ$ What it looks like: the flat orange "melt" step in the figure, climbed with $m = 0.150$ kg.

L1.2

Q: How much heat turns $80$ g of water at $100° C$ into steam at $100° C$ ?

Recall Solution

Both start and end are at $100° C$ → temperature does not change → vaporisation plateau, use $Q = m L_{v}$ . $Q = m L_{v} = 0.080 \times 22.6 \times 1 0^{5} = 1.808 \times 1 0^{5} J \approx 181 kJ$

L1.3

Q: A student adds $Q = 3.34 \times 1 0^{4}$ J to a chunk of ice at $0° C$ and it all melts with none left over. What was the mass of the ice?

Recall Solution

Rearrange the plateau formula $Q = m L_{f}$ for mass: $m = \frac{Q}{L _{f}} = \frac{3.34 \times 1 0 ^{4}}{3.34 \times 1 0 ^{5}} = 0.100 kg = 100 g$

Level 2 — Application

Goal: two or more terms in one calculation; split the journey at every kink.

L2.1

Q: Heat $200$ g of ice from $- 15° C$ up to water at $0° C$ (i.e. warm it, then melt it).

Recall Solution

Two pieces: slope (warm ice) then plateau (melt).

Warm ice $- 15 \to 0$ : $Δ T = 15$ K, $Q_{1} = m c_{ice} Δ T = 0.200 \times 2100 \times 15 = 6300 J$
Melt at $0° C$ : $Q_{2} = m L_{f} = 0.200 \times 3.34 \times 1 0^{5} = 66800 J$
Total: $Q = Q_{1} + Q_{2} = 6300 + 66800 = 73100 J = 73.1 kJ$ What it looks like: the first grey slope + the first orange plateau of the staircase.

L2.2

Q: How much heat raises $250$ g of water from $20° C$ all the way to steam at $100° C$ ?

Recall Solution

Slope (warm water $20 \to 100$ ) then plateau (boil).

Warm water: $Δ T = 80$ K, $Q_{1} = m c_{w} Δ T = 0.250 \times 4186 \times 80 = 83720 J$
Boil at $100° C$ : $Q_{2} = m L_{v} = 0.250 \times 22.6 \times 1 0^{5} = 565000 J$
Total: $Q = 83720 + 565000 = 648720 J \approx 649 kJ$ Notice the boil term dwarfs the warming term — $L_{v}$ is vast.

L2.3

Q: Steam at $100° C$ , mass $30$ g, is condensed and the resulting water is then cooled to $40° C$ . How much heat is released?

Recall Solution

Going "downhill" releases heat, but the magnitudes use the same formulas.

Condense at $100° C$ (plateau, releases $L_{v}$ ): $Q_{1} = m L_{v} = 0.030 \times 22.6 \times 1 0^{5} = 67800 J$
Cool water $100 \to 40$ : $Δ T = 60$ K, $Q_{2} = m c_{w} Δ T = 0.030 \times 4186 \times 60 = 7534.8 J$
Total released: $Q = 67800 + 7534.8 = 75334.8 J \approx 75.3 kJ$

Level 3 — Analysis

Goal: energy conservation across a mixture (method of mixtures) — heat lost = heat gained.

L3.1

Q: $400$ g of water at $30° C$ is cooled by dropping in ice at $0° C$ , and it settles at $0° C$ with some ice fully melted. What mass of ice melted?

Recall Solution

Energy conservation: heat lost by the warm water = heat gained (as latent heat) by the melting ice.

Heat released by water cooling $30 \to 0$ : $Q = m_{w} c_{w} Δ T = 0.400 \times 4186 \times 30 = 50232 J$
This melts ice via $Q = m_{ice} L_{f}$ : $m_{ice} = \frac{Q}{L _{f}} = \frac{50232}{3.34 \times 1 0 ^{5}} \approx 0.1504 kg \approx 150 g$

L3.2

Q: A calorimeter holds $250$ g of water at $60° C$ . You add $50$ g of ice at $0° C$ . Assuming all the ice melts, find the final temperature (ignore the calorimeter's own heat capacity).

Recall Solution

Let the final temperature be $T$ . Two things gain heat (the ice melts, then that meltwater warms $0 \to T$ ); one thing loses heat (the original water cools $60 \to T$ ). Set gained = lost.

Heat gained by the ice side: $Q_{gain} = melt m_{i} L_{f} + warm meltwater m_{i} c_{w} (T - 0)$ Heat lost by the warm water: $Q_{lost} = m_{w} c_{w} (60 - T)$ Set equal: $m_{i} L_{f} + m_{i} c_{w} T = m_{w} c_{w} (60 - T)$ Numbers ( $m_{i} = 0.05$ , $m_{w} = 0.25$ ): $(0.05) (3.34 \times 1 0^{5}) + (0.05) (4186) T = (0.25) (4186) (60 - T)$ $16700 + 209.3 T = 62790 - 1046.5 T$ $1255.8 T = 46090 \Rightarrow T = 36.7° C$ Sanity check: $T$ lies between $0$ and $60° C$ , so our "all ice melts" assumption is consistent.

Level 4 — Synthesis

Goal: full heating curves and multi-region journeys with sign/direction bookkeeping.

L4.1

Q: Heat $40$ g of ice at $- 20° C$ all the way to steam at $120° C$ . Give the total heat and state which single step dominates.

Recall Solution

Five regions — slope, plateau, slope, plateau, slope ( $m = 0.040$ kg):

Step	Formula	Value (J)
Warm ice $- 20 \to 0$ ( $Δ T = 20$ )	$m c_{ice} (20)$	$1680$
Melt at $0° C$	$m L_{f}$	$13360$
Warm water $0 \to 100$ ( $Δ T = 100$ )	$m c_{w} (100)$	$16744$
Boil at $100° C$	$m L_{v}$	$90400$
Warm steam $100 \to 120$ ( $Δ T = 20$ )	$m c_{steam} (20)$	$1608$

$Q_{total} = 1680 + 13360 + 16744 + 90400 + 1608 = 123792 J \approx 1.238 \times 1 0^{5} J$ Dominant step: the boiling plateau ( $90.4$ kJ, ~73 % of the total) because $L_{v}$ is by far the largest quantity.

L4.2

Q: You supply a fixed $50, 000$ J of heat to $60$ g of ice starting at $- 10° C$ . What is the final state and temperature?

Recall Solution

Spend the energy along the staircase and see how far $50000$ J takes us ( $m = 0.060$ ).

Warm ice $- 10 \to 0$ : needs $m c_{ice} (10) = 0.060 \times 2100 \times 10 = 1260$ J. Remaining: $50000 - 1260 = 48740$ J.
Melt all the ice: needs $m L_{f} = 0.060 \times 3.34 \times 1 0^{5} = 20040$ J. Remaining: $48740 - 20040 = 28700$ J.
Warm the water toward $100° C$ : full climb would need $m c_{w} (100) = 0.060 \times 4186 \times 100 = 25116$ J. We have $28700$ J — more than enough, water reaches $100° C$ . Remaining: $28700 - 25116 = 3584$ J.
Boil some water: fully boiling all $60$ g needs $m L_{v} = 135600$ J, but only $3584$ J is left. So only part boils: $m_{boiled} = \frac{3584}{22.6 \times 1 0 ^{5}} = 1.586 \times 1 0^{- 3} kg \approx 1.59 g$ Final state: water and steam coexist at $100° C$ — about $1.59$ g has boiled off, leaving $\approx 58.4$ g of liquid water. The temperature is stuck at $100° C$ (we ran out of energy mid-plateau).

Level 5 — Mastery

Goal: partial phase changes, equilibrium where not everything melts, and reasoning about which side "wins."

L5.1

Q: $100$ g of ice at $0° C$ is added to $60$ g of water at $50° C$ . Does all the ice melt? If not, find the final temperature and the mass of ice remaining.

Recall Solution

First test the assumption "all ice melts" by comparing energy available vs energy required.

Heat the warm water can give up cooling to $0° C$ : $Q_{available} = m_{w} c_{w} Δ T = 0.060 \times 4186 \times 50 = 12558 J$
Heat needed to melt all the ice: $Q_{needed} = m_{i} L_{f} = 0.100 \times 3.34 \times 1 0^{5} = 33400 J$ Since $12558 < 33400$ , not all the ice melts. The system settles at $0° C$ (ice + water coexist).
Mass of ice actually melted by the available $12558$ J: $m_{melt} = \frac{Q _{available}}{L _{f}} = \frac{12558}{3.34 \times 1 0 ^{5}} = 0.0376 kg = 37.6 g$
Ice remaining: $100 - 37.6 = 62.4 g of ice, final temperature 0° C$

L5.2

Q: How much steam at $100° C$ must condense onto $500$ g of water at $20° C$ to raise it to $80° C$ ? (The condensed steam becomes water and also cools to $80° C$ .)

Recall Solution

Let $m_{s}$ = mass of steam. It releases heat in two stages (condense, then cool $100 \to 80$ ); the cold water absorbs it warming $20 \to 80$ .

Heat gained by the cold water: $Q_{gain} = m_{w} c_{w} Δ T = 0.500 \times 4186 \times 60 = 125580 J$
Heat released by the steam: $Q_{rel} = m_{s} L_{v} + m_{s} c_{w} (100 - 80) = m_{s} (22.6 \times 1 0^{5}) + m_{s} (4186) (20)$ $= m_{s} (2.26 \times 1 0^{6} + 83720) = m_{s} (2.34372 \times 1 0^{6})$
Set gained = released: $m_{s} = \frac{125580}{2.34372 \times 1 0 ^{6}} = 0.05359 kg \approx 53.6 g$ Insight: because condensing steam dumps its enormous $L_{v}$ , a small mass of steam heats a large mass of water — this is exactly why a steam burn is so much worse than a boiling-water burn.

Connections

Specific heat capacity — every sloped step above.
Calorimetry — method of mixtures — the energy-balance backbone of L3–L5.
Kinetic theory of gases — why the plateaus sit at constant temperature.
Evaporation vs boiling — the steam-burn intuition of L5.2.
First law of thermodynamics — where the $L_{v}$ energy actually goes.
Phase diagrams — the $P$ – $T$ picture of these same transitions.