Principle of Calorimetry & Mixtures — NEET Physics MCQs with Solutions
Free NEET Physics Principle of Calorimetry & Mixtures 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 — Principle of Calorimetry & Mixtures · easy · theory
The principle of calorimetry states that, for an isolated system:
A. Heat lost by the hot body = heat gained by the cold body ✓ Correct
B. Heat always flows from cold to hot
C. The final temperature is the average of the initial temperatures
D. No heat is exchanged
Solution: Core equation: Heat lost = Heat gained (conservation of energy, with no losses to the surroundings).
Q2 — Principle of Calorimetry & Mixtures · easy · theory
The principle of calorimetry is a direct consequence of the law of:
A. Conservation of momentum
B. Conservation of energy ✓ Correct
C. Conservation of mass
D. Inertia
Solution: Heat lost = heat gained is simply energy conservation applied to thermal exchange.
Q3 — Principle of Calorimetry & Mixtures · easy · theory
When two liquids of EQUAL mass and EQUAL specific heat at temperatures T₁ and T₂ are mixed (no losses), the final temperature is:
A. Closer to the hotter one
B. √(T₁T₂)
C. (T₁ + T₂)/2 ✓ Correct
D. T₁ + T₂
Solution: Equal mc on both sides gives the simple mean. Speed-trick: any asymmetry (mass or c) pulls T_mix towards the larger mc side.
Q4 — Principle of Calorimetry & Mixtures · medium · theory
The general formula for the final temperature when two bodies (m₁,c₁,T₁) and (m₂,c₂,T₂) are mixed is:
A. $T_{mix} = \dfrac{T_1 + T_2}{2}$ always
B. $T_{mix} = T_1 - T_2$
C. $T_{mix} = \dfrac{m_1c_1T_1 + m_2c_2T_2}{m_1c_1 + m_2c_2}$ ✓ Correct
D. $T_{mix} = \dfrac{m_1T_1 + m_2T_2}{c_1 + c_2}$
Solution: Setting heat lost = heat gained gives the mc-weighted average of the temperatures.
Q5 — Principle of Calorimetry & Mixtures · medium · theory
The water equivalent of a calorimeter is:
A. The mass of water having the same heat capacity as the calorimeter (W = m_c·c_c/c_w) ✓ Correct
B. The heat it loses per second
C. Its mass in grams
D. The volume of water it can hold
Solution: W = m_c c_c / c_w: replace the calorimeter by W grams of water in the heat-balance equation.
Q6 — Principle of Calorimetry & Mixtures · easy · theory
The final temperature of a mixture always lies:
A. Between the initial temperatures of the two bodies ✓ Correct
B. Below both initial temperatures
C. Above both initial temperatures
D. Exactly at the mean always
Solution: Heat flows from hot to cold until a common intermediate temperature is reached; T_mix can never lie outside the range [T_cold, T_hot].
Q7 — Principle of Calorimetry & Mixtures · medium · theory
A calorimeter is usually made of a metal (e.g. copper) because metals have:
A. Low melting point
B. High density only
C. Very high specific heat
D. Low specific heat and high conductivity, so they absorb little heat and equilibrate fast ✓ Correct
Solution: A low specific heat means a small water equivalent (small correction), and high conductivity ensures quick temperature uniformity.
Q8 — Principle of Calorimetry & Mixtures · medium · theory
If equal masses of two DIFFERENT liquids (c₁ > c₂) at temperatures T₁ (hot, c₁) and T₂ (cold, c₂) are mixed, the final temperature lies:
A. Exactly midway
B. Closer to T₁ (the liquid with the larger mc dominates) ✓ Correct
C. Closer to T₂
D. At T₂ always
Solution: T_mix is the mc-weighted mean; the body with larger heat capacity (here the hot liquid with larger c) shifts the result towards its own temperature.
Q9 — Principle of Calorimetry & Mixtures · easy · numerical
100 g of water at 80°C is mixed with 100 g of water at 20°C. The final temperature is:
A. 60°C
B. 40°C
C. 50°C ✓ Correct
D. 100°C
Solution: Equal masses of the same liquid: T_mix = (80 + 20)/2 = 50°C.
Q10 — Principle of Calorimetry & Mixtures · easy · numerical
200 g of water at 90°C is mixed with 100 g of water at 30°C. The final temperature is:
A. 50°C
B. 70°C ✓ Correct
C. 80°C
D. 60°C
Solution: T = (200×90 + 100×30)/(200 + 100) = (18000 + 3000)/300 = 70°C.
Q11 — Principle of Calorimetry & Mixtures · easy · numerical
50 g of water at 100°C is added to 200 g of water at 25°C. The final temperature is:
A. 35°C
B. 50°C
C. 40°C ✓ Correct
D. 62.5°C
Solution: T = (50×100 + 200×25)/250 = (5000 + 5000)/250 = 40°C.
Q12 — Principle of Calorimetry & Mixtures · easy · numerical
A 100 g copper calorimeter (c = 0.1 cal/g°C) has a water equivalent of:
A. 10 g ✓ Correct
B. 1 g
C. 1000 g
D. 100 g
Solution: W = mc/c_w = 100 × 0.1/1 = 10 g of water.
Q13 — Principle of Calorimetry & Mixtures · medium · numerical
A calorimeter of water equivalent 10 g contains 90 g of water at 20°C. When 100 g of water at 60°C is poured in, the final temperature is:
A. 45°C
B. 35°C
C. 40°C ✓ Correct
D. 50°C
Solution: Effective cold mass = 90 + 10 = 100 g. T = (100×60 + 100×20)/200 = 40°C. Speed-trick: add W to the water mass and proceed normally.
Q14 — Principle of Calorimetry & Mixtures · medium · numerical
A 100 g metal block at 100°C is dropped into 200 g of water at 25°C; the final temperature is 30°C. The specific heat of the metal is:
A. ≈ 0.143 cal/g°C ✓ Correct
B. ≈ 1 cal/g°C
C. ≈ 0.5 cal/g°C
D. ≈ 0.07 cal/g°C
Solution: Heat lost = heat gained: 100c(100−30) = 200×1×(30−25) ⇒ 7000c = 1000 ⇒ c ≈ 0.143 cal/g°C.
Q15 — Principle of Calorimetry & Mixtures · medium · numerical
300 g of a liquid (c = 0.5 cal/g°C) at 80°C is mixed with 150 g of water at 20°C. The final temperature is:
A. 50°C ✓ Correct
B. 35°C
C. 40°C
D. 60°C
Solution: mc(hot) = 300×0.5 = 150; mc(cold) = 150×1 = 150 — equal! T = (80 + 20)/2 = 50°C. Speed-trick: equal mc ⇒ simple mean, whatever the masses.
Q16 — Principle of Calorimetry & Mixtures · medium · numerical
Three equal masses of water at 20°C, 40°C and 60°C are mixed. The final temperature is:
A. 40°C ✓ Correct
B. 30°C
C. 50°C
D. 45°C
Solution: Equal masses, same c: T = (20 + 40 + 60)/3 = 40°C.
Q17 — Principle of Calorimetry & Mixtures · medium · numerical
Equal masses of water at 60°C and a liquid of specific heat 0.5 cal/g°C at 20°C are mixed. The final temperature is:
A. ≈ 46.7°C ✓ Correct
B. 40°C
C. 30°C
D. 50°C
Solution: T = (1×60 + 0.5×20)/(1 + 0.5) = (60 + 10)/1.5 ≈ 46.7°C — closer to the water's 60°C because water has the larger mc.
Q18 — Principle of Calorimetry & Mixtures · medium · numerical
When 100 g of hot milk at 80°C is poured into a 40 g glass cup (c = 0.2 cal/g°C, take c_milk = 1 cal/g°C) at 20°C, the final temperature is:
A. 50°C
B. ≈ 70°C
C. ≈ 60°C
D. ≈ 75.6°C ✓ Correct
Solution: Cup's water equivalent = 40×0.2 = 8 g. T = (100×80 + 8×20)/108 = (8000 + 160)/108 ≈ 75.6°C.
Q19 — Principle of Calorimetry & Mixtures · medium · numerical
A calorimeter contains 70 g of water at 20°C. When 30 g of water at 90°C is added, the final temperature is 40°C. The water equivalent of the calorimeter is:
A. 15 g
B. 10 g
C. 2 g
D. 5 g ✓ Correct
Solution: Heat lost = 30(90−40) = 1500 cal. Heat gained = (70 + W)(40−20) = (70+W)20. So 70 + W = 75 ⇒ W = 5 g.
Q20 — Principle of Calorimetry & Mixtures · medium · numerical
10 g of water at 70°C is mixed with 5 g of water at 10°C. The final temperature is:
A. 45°C
B. 50°C ✓ Correct
C. 60°C
D. 40°C
Solution: T = (10×70 + 5×10)/15 = (700 + 50)/15 = 50°C.