Ice-Water-Steam Mixing Problems — NEET Physics MCQs with Solutions
Free NEET Physics Ice-Water-Steam Mixing Problems MCQs with step-by-step solutions (20 questions). Part of Calorimetry & Thermal Properties. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Ice-Water-Steam Mixing Problems · medium · theory
When ice at 0°C is mixed with water, before assuming a final temperature one must FIRST check:
A. The volume of the container
B. Whether the heat available from the water can melt all the ice ✓ Correct
C. The colour of the ice
D. The atmospheric pressure only
Solution: NEET speed-check: compare heat available (m_w c ΔT down to 0°C) with heat needed (m_ice L_f). If available < needed, only part of the ice melts and T_final = 0°C.
Q2 — Ice-Water-Steam Mixing Problems · medium · theory
If, on mixing ice and water, the heat released by the water in cooling to 0°C is LESS than the heat needed to melt all the ice, the final state is:
A. Steam
B. All ice below 0°C
C. All water above 0°C
D. An ice-water mixture at exactly 0°C ✓ Correct
Solution: The system settles at 0°C with some ice unmelted — a mixture at 0°C. Pitfall: never report a final temperature below the coldest or above the hottest component.
Q3 — Ice-Water-Steam Mixing Problems · easy · theory
The heat-flow sequence for converting ice at −10°C to steam at 100°C is:
A. Melt → warm ice → vaporize → warm water
B. Vaporize → melt → warm
C. A single continuous temperature rise
D. Warm ice → melt at 0°C → warm water → vaporize at 100°C ✓ Correct
Solution: Four stages: m·c_i·(10) + m·L_f + m·c_w·(100) + m·L_v — two warmings and two flat phase changes on the heating curve.
Q4 — Ice-Water-Steam Mixing Problems · easy · theory
On a temperature-vs-heat (heating) curve of ice being converted to steam, the flat (horizontal) portions represent:
A. Phase changes at constant temperature (melting and boiling) ✓ Correct
B. Cooling stages
C. Rapid temperature rise
D. Loss of mass
Solution: Flat segments = latent-heat stages (0°C melting, 100°C boiling); sloped segments = specific-heat warming. The boiling plateau is much longer (540 vs 80 cal/g).
Q5 — Ice-Water-Steam Mixing Problems · medium · theory
Equal masses of ice at 0°C and water at 80°C are mixed. Without calculation, the expected final state is:
A. All water at 80°C
B. All water at exactly 0°C (heat available exactly melts the ice) ✓ Correct
C. All water at 40°C
D. Ice-water mixture at 0°C with ice left over
Solution: Water gives m×1×80 = 80m cal cooling to 0°C; melting needs m×80 = 80m cal — an exact match. All ice just melts, final T = 0°C. A NEET classic!
Q6 — Ice-Water-Steam Mixing Problems · medium · theory
When steam is passed into cold water, the final temperature is comparatively high because:
A. Water cannot absorb heat
B. Steam is above 100°C
C. Steam has a very high specific heat
D. Each gram of condensing steam releases 540 cal before it even starts cooling as water ✓ Correct
Solution: The huge latent heat of condensation (540 cal/g) dominates the heat balance, which is why small amounts of steam raise water temperature dramatically.
Q7 — Ice-Water-Steam Mixing Problems · easy · theory
The specific heat of ice is 0.5 cal/g°C. This means warming 1 g of ice from −10°C to 0°C requires ______ compared to warming 1 g of water by the same 10°C:
A. Half the heat (5 cal vs 10 cal) ✓ Correct
B. Twice the heat
C. No heat
D. The same heat
Solution: Ice: 1 × 0.5 × 10 = 5 cal; water: 1 × 1 × 10 = 10 cal. Ice (and steam) have half the specific heat of water.
Q8 — Ice-Water-Steam Mixing Problems · medium · theory
If the final mixture of ice and water settles at 0°C with ice remaining, the amount of ice melted is found from:
A. m_melted = (heat × c_w)
B. m_melted = 0 always
C. m_melted = total ice mass always
D. m_melted = (heat released by water cooling to 0°C)/L_f ✓ Correct
Solution: All the available heat goes into melting: m_melted = Q_available/80 (in cal and grams). The rest of the ice stays frozen.
Q9 — Ice-Water-Steam Mixing Problems · medium · numerical
10 g of ice at 0°C is mixed with 100 g of water at 40°C. The final temperature is:
A. 0°C
B. 20°C
C. ≈ 29.1°C ✓ Correct
D. ≈ 36.4°C
Solution: Check: water can give 100×40 = 4000 cal; melting needs 10×80 = 800 cal — plenty. Balance: 100(40−T) = 800 + 10(T−0) ⇒ 4000 − 100T = 800 + 10T ⇒ T = 3200/110 ≈ 29.1°C.
Q10 — Ice-Water-Steam Mixing Problems · medium · numerical
50 g of ice at 0°C is added to 50 g of water at 40°C. The final state is:
A. All water at 0°C
B. All water at 20°C
C. 0°C with all ice melted exactly
D. 0°C with 25 g of ice still unmelted ✓ Correct
Solution: Available: 50×40 = 2000 cal. Needed to melt all: 50×80 = 4000 cal. Not enough ⇒ T = 0°C; melted = 2000/80 = 25 g, so 25 g of ice remains.
Q11 — Ice-Water-Steam Mixing Problems · easy · numerical
The heat required to convert 10 g of ice at −10°C into water at 0°C is:
A. 800 cal
B. 1600 cal
C. 900 cal
D. 850 cal ✓ Correct
Solution: Warm ice: 10 × 0.5 × 10 = 50 cal; melt: 10 × 80 = 800 cal. Total = 850 cal.
Q12 — Ice-Water-Steam Mixing Problems · medium · numerical
The total heat needed to convert 5 g of ice at −20°C into steam at 100°C is:
A. 2700 cal
B. 3650 cal ✓ Correct
C. 3100 cal
D. 3600 cal
Solution: Warm ice: 5×0.5×20 = 50; melt: 5×80 = 400; warm water: 5×1×100 = 500; vaporize: 5×540 = 2700. Total = 3650 cal.
Q13 — Ice-Water-Steam Mixing Problems · medium · numerical
1 g of steam at 100°C is passed into 20 g of water at 20°C. The final temperature is (approximately):
A. ≈ 25°C
B. ≈ 49.5°C ✓ Correct
C. ≈ 60°C
D. 100°C
Solution: Steam gives: 1×540 + 1×(100−T); water takes: 20(T−20). 540 + 100 − T = 20T − 400 ⇒ 1040 = 21T ⇒ T ≈ 49.5°C. Note how just 1 g of steam heats 20 g of water by ~30°C!
Q14 — Ice-Water-Steam Mixing Problems · medium · numerical
10 g of steam at 100°C is mixed with 90 g of water at 0°C. The final temperature is:
A. 54°C
B. 100°C
C. 36°C
D. 64°C ✓ Correct
Solution: Steam: 10×540 + 10(100−T); water: 90(T−0). 5400 + 1000 − 10T = 90T ⇒ 6400 = 100T ⇒ T = 64°C.
Q15 — Ice-Water-Steam Mixing Problems · medium · numerical
80 g of ice at 0°C is mixed with 80 g of water at 80°C. The final state is:
A. All water at 10°C
B. 40°C water
C. All water at 0°C (exact melt) ✓ Correct
D. Ice-water mixture with 40 g ice left
Solution: Available: 80×80 = 6400 cal; needed: 80×80 = 6400 cal — exact. All ice melts and T_final = 0°C.
Q16 — Ice-Water-Steam Mixing Problems · medium · numerical
20 g of ice at 0°C is dropped into 60 g of water at 60°C. The final temperature is:
A. 25°C ✓ Correct
B. 30°C
C. 45°C
D. 0°C
Solution: Check: 60×60 = 3600 > 20×80 = 1600 ⇒ all melts. Balance: 60(60−T) = 1600 + 20T ⇒ 3600 − 60T = 1600 + 20T ⇒ T = 2000/80 = 25°C.
Q17 — Ice-Water-Steam Mixing Problems · medium · numerical
The mass of steam at 100°C needed to just melt 27 g of ice at 0°C (final state: all water at 0°C… steam condenses and cools 100°C) is:
A. ≈ 3.4 g ✓ Correct
B. 27 g
C. ≈ 5 g
D. ≈ 1 g
Solution: Ice needs 27×80 = 2160 cal. Each gram of steam gives 540 + 100 = 640 cal (condense + cool to 0°C). m = 2160/640 ≈ 3.4 g. Speed-trick: 1 g of steam ≈ melts 8 g of ice.
Q18 — Ice-Water-Steam Mixing Problems · medium · numerical
100 g of ice at −10°C is mixed with 100 g of water at 10°C. The final state is:
A. All ice at 0°C
B. All water at 0°C
C. Water at 5°C
D. 0°C with an ice-water mixture (only ~6.25 g of ice melts) ✓ Correct
Solution: Water gives 100×10 = 1000 cal cooling to 0°C. Warming the ice to 0°C needs 100×0.5×10 = 500 cal, leaving 500 cal, which melts only 500/80 ≈ 6.25 g. Mixture at 0°C.
Q19 — Ice-Water-Steam Mixing Problems · easy · numerical
The heat released when 10 g of steam at 100°C is converted into water at 20°C is:
A. 7200 cal
B. 6200 cal ✓ Correct
C. 800 cal
D. 5400 cal
Solution: Condense: 10×540 = 5400; cool: 10×1×80 = 800. Total released = 6200 cal.
Q20 — Ice-Water-Steam Mixing Problems · medium · numerical
A heating curve shows heat on the x-axis for 1 g of ice starting at 0°C. The length of the melting plateau compared with the boiling plateau is in the ratio:
A. 1 : 1
B. 80 : 540 = 4 : 27 ✓ Correct
C. 100 : 540
D. 540 : 80
Solution: Plateau lengths are proportional to latent heats: fusion 80 cal vs vaporization 540 cal ⇒ 4 : 27 — the boiling plateau is ~6.75× longer.