PhysicsICSE Class 10 · Latent Heat

Mixtures of Ice and Water: Heat Balance, Final Temperature and Mass of Ice (ICSE Class 10)

Updated 16 October 2026 · Fact-checked

When ice at 0 °C melts completely in warm water, heat lost by the water = heat to melt the ice + heat to warm the melted ice: m_w c(t_w − t) = m_i L + m_i c t. If the water cannot supply enough heat, only part of the ice melts and the final temperature is 0 °C.

Ice first melts at a steady 0 °C (m L), then the melted water warms up (m c t); the warm water must supply both amounts of heat.

Illustration drawn for QuestionBuddy.

Understand Mixtures of Ice and Water

The idea. This is the principle of method of mixtures with one extra step: the ice must melt before its water can warm up. With no heat lost to the surroundings:

heat lost by the warm water (and calorimeter) = heat gained by the ice.

Heat gained by ice, stage by stage. For ice of mass $m_{i}$ ending as water at temperature $t$ :
1. If the ice starts below 0 °C, at $t_{i}$ : warm it to 0 °C: $m_{i} c_{i ce} (0 - t_{i})$ .
2. Melt it at 0 °C: $m_{i} L$ .
3. Warm the melted ice (now water) from 0 °C to $t$ : $m_{i} c_{w a t er} (t - 0)$ .

Step 3 is the one most often forgotten. The melted ice is water at 0 °C; it too must reach the final temperature.

The standard equation (ice at 0 °C, all of it melts):

$m_{w} c_{w} (t_{w} - t) = m_{i} L + m_{i} c_{w} t .$

With a calorimeter (mass $m_{c}$ , specific heat $c_{c}$ ) that starts at $t_{w}$ , add $m_{c} c_{c} (t_{w} - t)$ to the left side.

Step 0: does all the ice melt? Before solving for $t$ , compare:

heat the warm water can give out in cooling to 0 °C: $m_{w} c_{w} t_{w}$ ;
heat needed to bring all the ice to 0 °C and melt it: $m_{i} c_{i ce} (0 - t_{i}) + m_{i} L$ .

If the first is larger, all the ice melts and the final temperature is above 0 °C: solve the full equation. If it is smaller, only part of the ice melts and the final temperature is 0 °C. The mass that melts is

$m_{m e l t e d} = \frac{heat available - heat to bring ice to 0 ° C}{L} .$

Skipping this check can give a final temperature below 0 °C, which is impossible for a mixture of warm water and ice at 0 °C.

Typical question types.

Find	Unknown in the equation
Mass of ice needed to cool water to $t$	$m_{i}$
Final temperature	$t$
Mass of ice that melts (not enough heat)	$m_{m e l t e d}$ , with $t = 0$ °C
Specific latent heat (experiment)	$L$

**Finding $L$ by the method of mixtures.** Warm water of known mass and temperature is taken in a calorimeter. Small pieces of dry ice at 0 °C are added and stirred until they melt, and the lowest steady temperature is noted. The calorimeter is weighed again: the increase in mass is the mass of ice added. Then

$L = \frac{( m _{w} c _{w} + m _{c} c _{c} ) ( t _{w} - t ) - m _{i} c _{w} t}{m _{i}} .$

The ice is dried (wiped) first, because water clinging to it would add mass without needing latent heat, giving too low a value of $L$ .

Why ice cools better than ice-cold water. Each gram of ice at 0 °C absorbs 336 J just to melt, before it begins to warm. So a drink cooled with ice ends up much colder than one cooled with the same mass of water at 0 °C.

Mixtures of Ice and Water: formulas and key facts

Ice at 0 °C melting completely: $m_{w} c_{w} (t_{w} - t) = m_{i} L + m_{i} c_{w} t$; t is the final temperature.
Ice below 0 °C: $m_{w} c_{w} (t_{w} - t) = m_{i} c_{i ce} (0 - t_{i}) + m_{i} L + m_{i} c_{w} t$; Three stages for the ice.
Check: does all the ice melt?: $m_{w} c_{w} t_{w} vs m_{i} c_{i ce} (0 - t_{i}) + m_{i} L$; If the left is smaller, t = 0 °C.
Values: $L = 336 J g^{- 1} ∣ c_{w} a t er = 4.2 J g^{- 1} ° C^{- 1} ∣ c_{i} ce = 2.1 J g^{- 1} ° C^{- 1}$; Use g with J g⁻¹ throughout.

How to solve Mixtures of Ice and Water questions

Sketch the heating curve for the ice, write one term per stage, and check that there is enough heat before solving.

1List the warm bodies (water, calorimeter) and the ice, with masses and starting temperatures.
2Check whether all the ice can melt: compare $m_{w} c_{w} t_{w}$ (plus calorimeter) with the heat to bring the ice to 0 °C and melt it.
3If all melts, write: heat lost by warm bodies $=$ (warm ice to 0 °C) $+ m_{i} L + m_{i} c_{w} t$ .
4If not, the final temperature is 0 °C; find the mass melted from the heat available.
5Keep units consistent (g with J g⁻¹ and J g⁻¹ °C⁻¹); use the values of $L$ and $c$ given.
6Check the answer: $t$ must lie between 0 °C and the warm water's temperature.

Mixtures of Ice and Water in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Ice for a party punch

Suppose you have 2 kg of fruit punch at 30 °C and want it at 10 °C, using ice cubes at 0 °C. Treat the punch as water and ignore heat from the room.

How much ice is needed?

Try it first, then show the solutionSolution

Heat lost by punch $= 2000 \times 4.2 \times 20 = 168000$ J.
Each gram of ice gains $336 + 4.2 \times 10 = 378$ J (melt, then warm to 10 °C).
$m = \frac{168 000}{378} \approx 444$ g.

Answer: About 444 g of ice.

Memory hook: Each gram of ice: 336 to melt, plus the warm-up.

Snow in a trekker's pot

Imagine a trekker adds 200 g of clean snow at 0 °C to 800 g of water at 80 °C in an insulated flask to get drinking water quickly.

What is the final temperature?

Try it first, then show the solutionSolution

Check: the water can give $800 \times 4.2 \times 80 = 268800$ J, far more than the $200 \times 336 = 67200$ J needed to melt the snow.
$800 \times 4.2 \times (80 - t) = 200 \times 336 + 200 \times 4.2 \times t$ .
$3360 (80 - t) = 67200 + 840 t \Rightarrow 201600 = 4200 t \Rightarrow t = 48$ °C.

Answer: 48 °C.

Memory hook: Melt first, then share the warmth.

Tricks to remember Mixtures of Ice and Water

Short memory aids. Say each one aloud once, then use it on a question the same day.

Melt, then warm
Ice gains mL first, THEN mcΔT for its melted water: never stop at mL.
Use it when: Writing the 'heat gained' side.
Check before you solve
'Enough heat?' first: if not, the answer is 0 °C.
Use it when: Avoiding impossible answers.
378 for 10 °C
One gram of ice at 0 °C ending at t °C gains (336 + 4.2t) J.
Use it when: Quick per-gram arithmetic.
Cancel the 4.2
336 = 80 × 4.2, so divide the whole equation by 4.2 to simplify.
Use it when: Faster calculation.

Mixtures of Ice and Water: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Heat gained by the ice · 1 mark · MCQBoard level

Ice of mass $m$ at 0 °C is added to warm water and all of it melts; the final temperature is $t$ . The heat gained by the ice is:
(a) $m L$
(b) $m L + m c t$
(c) $m c t$
(d) $m L - m c t$

Try it first, then show the solutionSolution

It melts ( $m L$ ) and then its water warms from 0 °C to $t$ ( $m c t$ ).

Answer: (b) $m L + m c t$

Example 2: Ice to cool a glass of water · 3 marks · Short answerBoard level

Find the mass of ice at 0 °C needed to cool 180 g of water from 55 °C to 10 °C. ( $L = 336$ J g⁻¹, $c = 4.2$ J g⁻¹ °C⁻¹)

Try it first, then show the solutionSolution

Heat lost by water $= 180 \times 4.2 \times 45 = 34020$ J. (1)
Heat gained by ice $= m \times 336 + m \times 4.2 \times 10 = 378 m$ . (1)
$378 m = 34020 \Rightarrow m = 90$ g. (1)

Answer: 90 g

Example 3: Finding the final temperature · 3 marks · Short answerHarder board level

30 g of crushed ice at 0 °C is stirred into 270 g of water at 45 °C in a vacuum flask. (a) Show that all the ice melts. (b) Find the final temperature. ( $L = 336$ J g⁻¹, $c = 4.2$ J g⁻¹ °C⁻¹)

Try it first, then show the solutionSolution

(a) Heat available to 0 °C $= 270 \times 4.2 \times 45 = 51030$ J; heat to melt the ice $= 30 \times 336 = 10080$ J. More than enough. (1)
(b) $270 \times 4.2 \times (45 - t) = 10080 + 30 \times 4.2 \times t$ , so $1134 (45 - t) = 10080 + 126 t$ . (1)
$51030 - 10080 = 1260 t \Rightarrow t = 32.5$ °C. (1)

Answer: (a) 51 030 J > 10 080 J (b) 32.5 °C

Example 4: Not enough heat · 3 marks · Short answerHarder board level

200 g of ice at 0 °C is added to 150 g of water at 32 °C in an insulated vessel. (a) Will all the ice melt? (b) Find the final temperature and the mass of ice left.

Try it first, then show the solutionSolution

(a) The water can give at most $150 \times 4.2 \times 32 = 20160$ J in cooling to 0 °C; melting all the ice needs $200 \times 336 = 67200$ J. No. (1)
(b) Final temperature $= 0$ °C. Ice melted $= \frac{20 160}{336} = 60$ g. (1)
Ice left $= 200 - 60 = 140$ g, floating in 210 g of water at 0 °C. (1)

Answer: (a) No (b) 0 °C; 140 g of ice left

Example 5: Ice from the freezer · 4 marks · Long answerHarder board level

50 g of ice at −8 °C is added to 260 g of water at 40 °C. Find the final temperature, assuming all the ice melts and no heat is lost. ( $c_{i ce} = 2.1$ , $c_{w a t er} = 4.2$ J g⁻¹ °C⁻¹, $L = 336$ J g⁻¹)

Try it first, then show the solutionSolution

Heat to warm ice to 0 °C $= 50 \times 2.1 \times 8 = 840$ J; to melt it $= 50 \times 336 = 16800$ J. (1)
Heat to warm melted ice to $t$ $= 50 \times 4.2 \times t = 210 t$ . (1)
Heat lost by water $= 260 \times 4.2 \times (40 - t) = 1092 (40 - t)$ . (1)
$43680 - 1092 t = 17640 + 210 t \Rightarrow 26040 = 1302 t \Rightarrow t = 20$ °C. (1)

Answer: 20 °C

Example 6: Measuring L in the laboratory · 3 marks · Short answerHOTS / competency

A copper calorimeter of mass 105 g ( $c = 0.4$ J g⁻¹ °C⁻¹) contains 190 g of water at 28 °C. Pieces of dried ice at 0 °C are added and stirred until the temperature falls to 10 °C and all the ice has melted. The calorimeter and contents now weigh 40 g more. Find the specific latent heat of fusion of ice.

Try it first, then show the solutionSolution

Heat lost by water and calorimeter $= (190 \times 4.2 + 105 \times 0.4) (28 - 10) = 840 \times 18 = 15120$ J. (1)
Heat gained by ice $= 40 L + 40 \times 4.2 \times 10 = 40 L + 1680$ . (1)
$40 L = 15120 - 1680 = 13440 \Rightarrow L = 336$ J g⁻¹. (1)

Answer: 336 J g⁻¹

Example 7: Case study: ice in lemonade · 4 marks · Case-basedHOTS / competency

A jug holds 400 g of lemonade at 30 °C (take $c = 4.2$ J g⁻¹ °C⁻¹). Four ice cubes of 25 g each, at 0 °C, are added. Ignore heat exchange with the jug and the room.
(a) Show that all the ice melts.
(b) Find the final temperature.
(c) Find the final temperature if 100 g of water at 0 °C were added instead of the ice.
(d) Explain the difference.

Try it first, then show the solutionSolution

(a) Available: $400 \times 4.2 \times 30 = 50400$ J; needed to melt 100 g: $33600$ J. Enough. (1)
(b) $1680 (30 - t) = 33600 + 420 t \Rightarrow 16800 = 2100 t \Rightarrow t = 8$ °C. (1)
(c) $1680 (30 - t) = 420 t \Rightarrow 50400 = 2100 t \Rightarrow t = 24$ °C. (1)
(d) The ice absorbs 33 600 J of latent heat just to melt, before it starts warming; the cold water absorbs heat only by warming. (1)

Answer: (a) 50 400 J > 33 600 J (b) 8 °C (c) 24 °C (d) Latent heat of fusion

Where students lose marks in Mixtures of Ice and Water

The slips examiners see most often, and what to write instead.

Mistake: Forgetting to warm the melted ice
After melting, the ice is water at 0 °C, which must also be warmed to the final temperature.
Wrong working
Heat lost by water = m × 336 only, for a final temperature of 5 °C.
Right working
Heat lost by water = m × 336 + m × 4.2 × 5.
Do this: Sketch the heating curve of the ice and write a term for each stage.
Usually costs: Reported by CISCE examiners for 2024 Q9(i).
Mistake: Wrong equation
Heat gained by the ice is mL + mcΔT, not mcΔT alone and not mL alone.
Wrong working
'mcΔt = mcΔt' or 'mcΔt = ml' for ice that melts and then warms.
Right working
'm_w c (t_w − t) = m_i L + m_i c t.'
Do this: Write the left and right sides in words before substituting.
Usually costs: Reported by CISCE examiners for 2025 Q9(i).
Mistake: Converting only some quantities to SI
Mass in g with c and L in J kg⁻¹ (or the reverse) makes the equation wrong even if the answer looks right.
Wrong working
Mass in grams, L = 3.36 × 10⁵ J kg⁻¹.
Right working
Mass in grams with L = 336 J g⁻¹ and c = 4.2 J g⁻¹ °C⁻¹.
Do this: Keep everything in the units the question gives.
Usually costs: Reported by CISCE examiners for 2025 Q9(i).
Mistake: Not checking whether all the ice melts
With too little warm water, the final temperature is 0 °C.
Wrong working
Solving the full equation and getting t = −6 °C.
Right working
Compare heat available with heat to melt; if less, t = 0 °C.
Do this: Make the check your first line.
Usually costs: Usually costs the answer marks.

Exam tips

Simplify by dividing through by 4.2: 336 = 80 × 4.2.
Write 'Heat lost by water = heat gained by ice' as the first line of your answer.
State 'assuming no heat is lost to the surroundings'.
Give the final temperature in °C and masses in g, with units.

Practice questions on Mixtures of Ice and Water

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Ice at 0 °C is placed in water at 0 °C in a perfectly insulated flask. After some time:
(a) all the ice melts
(b) all the water freezes
(c) nothing changes; both stay at 0 °C
(d) the temperature rises
Hint
Is there a temperature difference?
Answer
(c) nothing changes; both stay at 0 °C
Q2 · 1 mark · Very short answerBoard level
Write the heat-balance equation when ice of mass $m_{i}$ at 0 °C melts completely in water of mass $m_{w}$ at $t_{w}$ , reaching a final temperature $t$ .
Hint
Lost = melt + warm.
Answer
$m_{w} c (t_{w} - t) = m_{i} L + m_{i} c t$ .
Q3 · 2 marks · Short answerBoard level
Find the heat needed to turn 25 g of ice at 0 °C into water at 12 °C. ( $L = 336$ J g⁻¹, $c = 4.2$ J g⁻¹ °C⁻¹)
Hint
Two stages.
Answer
$25 \times 336 + 25 \times 4.2 \times 12 = 8400 + 1260 = 9660$ J.
Q4 · 2 marks · Short answerHarder board level
What mass of ice at 0 °C is needed to cool 210 g of water at 32 °C to 0 °C, with all the ice just melting?
Hint
No warming term: final temperature 0 °C.
Answer
$210 \times 4.2 \times 32 = m \times 336 \Rightarrow m = \frac{28 224}{336} = 84$ g.
Q5 · 3 marks · Short answerHarder board level
20 g of ice at 0 °C is added to 180 g of water at 40 °C. Find the final temperature.
Hint
Check first, then solve.
Answer
Enough heat (30 240 J available, 6720 J needed). $756 (40 - t) = 6720 + 84 t \Rightarrow 23520 = 840 t \Rightarrow t = 28$ °C.
Q6 · 3 marks · Short answerHarder board level
120 g of ice at 0 °C is added to 100 g of water at 60 °C. Does all the ice melt? Find the final temperature and the mass of ice that melts.
Hint
Compare 25 200 J with 40 320 J.
Answer
Water can give $100 \times 4.2 \times 60 = 25200$ J; melting all the ice needs $40320$ J. No: final temperature 0 °C; ice melted $= \frac{25 200}{336} = 75$ g.
Q7 · 2 marks · Short answerHarder board level
Find the heat needed to turn 30 g of ice at −10 °C into water at 0 °C. ( $c_{i ce} = 2.1$ J g⁻¹ °C⁻¹, $L = 336$ J g⁻¹)
Hint
Warm, then melt.
Answer
$30 \times 2.1 \times 10 + 30 \times 336 = 630 + 10080 = 10710$ J.
Q8 · 2 marks · Short answerHarder board level
Why should you check whether all the ice melts before solving for the final temperature?
Hint
What if the heat is not enough?
Answer
If the warm water cannot supply enough heat to melt all the ice, the final temperature is 0 °C and only part of the ice melts. Solving the full equation anyway would give an impossible temperature below 0 °C.
Q9 · 3 marks · Short answerHOTS / competency
300 g of water at 28 °C is in a copper calorimeter of mass 210 g ( $c = 0.4$ J g⁻¹ °C⁻¹). 40 g of ice at 0 °C is added. Find the final temperature.
Hint
Heat capacity of water + calorimeter = 1344 J °C⁻¹.
Answer
$1344 (28 - t) = 40 \times 336 + 40 \times 4.2 t \Rightarrow 37632 - 1344 t = 13440 + 168 t \Rightarrow 24192 = 1512 t \Rightarrow t = 16$ °C.
Q10 · 2 marks · Short answerHOTS / competency
In finding L by the method of mixtures, a student adds ice that is wet on the surface. Will the value of L found be too high or too low? Explain.
Hint
The extra mass needs no latent heat.
Answer
Too low. The clinging water is counted in the mass of 'ice' but needs no latent heat to melt, so the heat is shared over a larger mass than really melted, making L appear smaller.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Heat gained by m grams of ice at 0 °C ending at t °C?Show answer
m(336 + 4.2t) J.
2.Final temperature if not all the ice melts?Show answer
0 °C.
3.First step in an ice-mixture problem?Show answer
Check whether all the ice can melt.
4.Why dry the ice when finding L?Show answer
Clinging water would make L too low.
5.Better coolant: ice at 0 °C or water at 0 °C?Show answer
Ice.

Mixtures of Ice and Water: frequently asked questions

How do you find the mass of ice needed to cool water?

Set heat lost by the water equal to heat gained by the ice: m_w c(t_w − t) = m_i L + m_i c t, then solve for m_i. Include a term for warming the ice to 0 °C if it starts below 0 °C.

What happens if there is not enough warm water to melt all the ice?

Only part of the ice melts and the final temperature is 0 °C. The mass that melts equals the heat the water can give in cooling to 0 °C divided by L.

Why is the term m_i c t included?

After melting, the ice becomes water at 0 °C, which must also be warmed to the final temperature t; m_i c t is the heat it absorbs for this.