Heat lost by hot solid/liquid = heat gained by cold water + calorimeter. m_s·c_s·(T_s − T_f) = (m_w·c_w + W)·(T_f − T_w), where W is water-equivalent of calorimeter. Solve for c_s. Account for radiation losses by Newton's correction or by performing the experiment quickly.
-- NCERT Physics Lab Manual Class 11, Part 7, p. 119Experiment 09 Specific Heat Capacity
8 MCQs9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Experiment 09 Specific Heat Capacity, explained for NEET
The calorimeter is not a spectator. In the method of mixtures the vessel and the stirrer sit in the cold water, warm up with it, and absorb a real share of the heat the hot body gives up. A student who writes "heat lost by the solid = heat gained by the water" and stops there has already lost several per cent, always in the same direction — the specific heat comes out too small. The procedure in the NCERT Physics Lab Manual Class 11, Part 7, page 119 carries the calorimeter's contribution explicitly, as a water equivalent: the mass of water that would soak up the same heat for the same temperature rise. For a copper calorimeter plus stirrer of mass m and specific heat c, that water equivalent is simply W = mc in calorie-gram units.
The balance sheet is one line. Heat given out by the hot body as it falls from its initial temperature to the mixture temperature equals heat taken in by the cold water plus the calorimeter's water equivalent as both rise to that same mixture temperature. Everything else in the experiment exists to protect that line: the solid is heated in a steam chamber so its starting temperature is known, it is transferred dry and fast so no hot droplets ride in with it, the water is stirred so one thermometer reading really represents the whole mixture, and the calorimeter sits in an insulating jacket so the room does not join the accounting.
For the liquid version the unknown moves. A solid whose specific heat is already known becomes the heat source, and the same single equation is solved for the liquid instead. Nothing new is measured — only the unknown changes place.
Watch-out for the exam: whenever a question hands you a calorimeter's mass and its specific heat, or its water equivalent outright, that number belongs on the gain side. Leaving it out is the standard distractor and it is always an under-estimate.
Can you answer these Experiment 09 Specific Heat Capacity MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The SI unit of specific heat capacity is:
Show answer and why every option is right or wrong
Answer: C. Option C. Specific heat capacity is heat per unit mass per unit temperature rise, so its SI unit is J kg⁻¹ K⁻¹, as used throughout the calorimetry experiment in the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why A is wrong: A is wrong because J kg⁻¹ is energy per unit mass — a latent heat, not a specific heat; it carries no temperature in the denominator.
Why B is wrong: B is wrong because J K⁻¹ is the heat capacity of a whole object (the calorimeter, say), which depends on how much material is present; specific heat is per unit mass.
Why D is wrong: D is wrong because cal g⁻¹ is neither SI nor per-degree; the calorie-gram system is a working convenience in the lab record, and even there the specific heat carries a per-°C.
MCQ 2Easy RecallPractice
The principle on which the method of mixtures rests is best stated as:
Show answer and why every option is right or wrong
Answer: A. Option A. The experiment is an energy balance under an assumed adiabatic enclosure, and the calorimeter and stirrer are on the gain side along with the water — the statement used to set up the working equation in the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why B is wrong: B is wrong because the mixture temperature is a weighted mean, weighted by mass times specific heat; a plain average is right only in the special case of equal masses of the same substance.
Why C is wrong: C is wrong because flow stops at equal temperature, not equal heat content; a large cold mass at the same temperature holds far more internal energy than a small hot one did.
Why D is wrong: D is wrong because heat capacity is mass times specific heat, so masses of different substances cannot simply be added; this is the error that makes students drop the calorimeter term.
MCQ 3Easy RecallPractice
The water equivalent of a calorimeter is:
Show answer and why every option is right or wrong
Answer: D. Option D. Water equivalent replaces the calorimeter by an equal-behaving mass of water, so it can be added straight to the water mass on the gain side of the balance — the device used in the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why A is wrong: A is wrong because it describes the vessel's capacity in volume terms, which has nothing to do with how much heat the copper absorbs.
Why B is wrong: B is wrong because that defines the calorimeter's heat capacity. Water equivalent is a mass, numerically equal to the heat capacity only in the calorie-gram system where the specific heat of water is 1.
Why C is wrong: C is wrong because water equivalent is fixed by the calorimeter alone — its mass and its specific heat — and does not change when a different amount of water is poured in.
MCQ 4Direct ApplicationPractice
A calorimeter of water equivalent 2.0 × 10¹ g contains 1.60 × 10² g of water at 20.0 °C. A solid of mass 1.00 × 10² g at 100.0 °C is dropped in, and the mixture settles at 28.0 °C. The specific heat capacity of the solid, in cal g⁻¹ °C⁻¹, is:
Show answer and why every option is right or wrong
Answer: B. Option B. Heat gained = (160 + 20) × 1 × 8.0 = 1.44 × 10³ cal; heat lost = 100 × c × 72.0, so c = 0.20 cal g⁻¹ °C⁻¹ — the standard solid calculation of the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why A is wrong: A is wrong because it drops the calorimeter's water equivalent and also takes the solid's fall as 80 °C instead of 72 °C: 160 × 8.0 / (100 × 80.0) = 0.16.
Why C is wrong: C is wrong because it omits the calorimeter's 2.0 × 10¹ g water equivalent from the gain side: 160 × 8.0 / (100 × 72.0) = 0.18. The omission always lowers the result.
Why D is wrong: D is wrong because it inverts the ratio, dividing the solid's heat-capacity term by the heat gained; the answer would then be dimensionally an inverse specific heat.
MCQ 5Direct ApplicationPractice
In a calorimeter of negligible water equivalent, 1.00 × 10² g of water at 80.0 °C is mixed with 3.00 × 10² g of water at 20.0 °C. Neglecting losses to the surroundings, the temperature of the mixture is:
Show answer and why every option is right or wrong
Answer: D. Option D. With one substance the balance is m₁(80.0 − T) = m₂(T − 20.0): 100(80.0 − T) = 300(T − 20.0) gives T = 35.0 °C — the mass-weighted mean used as the check calculation in the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why A is wrong: A is wrong because 50.0 °C is the plain average of 80.0 and 20.0, which would be right only for equal masses; here the cold water outweighs the hot three to one.
Why B is wrong: B is wrong because it weights the result towards the cold water too heavily — it corresponds to a 1 : 5 mass ratio, not 1 : 3.
Why C is wrong: C is wrong because it weights towards the hot water, reversing which mass is larger; the hot portion is the smaller one here.
MCQ 6Direct ApplicationPractice
A copper calorimeter together with its stirrer has a mass of 8.0 × 10¹ g, and the specific heat capacity of copper is 0.09 cal g⁻¹ °C⁻¹. Its water equivalent is:
Show answer and why every option is right or wrong
Answer: A. Option A. Water equivalent W = mc = 80 × 0.09 = 7.2 g, the mass of water that absorbs the same heat per degree — the quantity carried into the gain side in the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why B is wrong: B is wrong because it uses the calorimeter's own mass, which would be its water equivalent only if copper had the same specific heat as water — it is about eleven times smaller.
Why C is wrong: C is wrong because it divides by the specific heat instead of multiplying: 80 / 0.09 = 8.9 × 10². Water equivalent must be smaller than the copper mass, never larger.
Why D is wrong: D is wrong because it quotes the specific heat itself with a mass unit attached; the mass of the calorimeter has been left out entirely.
MCQ 7CalculationPractice
A solid of mass 5.0 × 10¹ g and known specific heat capacity 0.40 cal g⁻¹ °C⁻¹ is heated to 100.0 °C and dropped into 5.00 × 10² g of a liquid at 25.0 °C held in a calorimeter of water equivalent 3.0 × 10¹ g. The mixture settles at 30.0 °C. The specific heat capacity of the liquid, in cal g⁻¹ °C⁻¹, is:
Show answer and why every option is right or wrong
Answer: C. Option C. Heat lost = 50 × 0.40 × 70.0 = 1.4 × 10³ cal; heat gained = (500 c + 30 × 1) × 5.0, so 2500 c + 150 = 1400 and c = 0.50 cal g⁻¹ °C⁻¹ — the liquid variant of the method described in the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why A is wrong: A is wrong because it leaves the calorimeter's 3.0 × 10¹ g water equivalent off the gain side: 1400 / (500 × 5.0) = 0.56, an over-estimate of the liquid's specific heat.
Why B is wrong: B is wrong because it simply repeats the solid's known specific heat; the two substances are unrelated, and the solid's value is an input, not the answer.
Why D is wrong: D is wrong because it takes the solid's fall as 100.0 → 25.0 °C, the liquid's starting temperature, instead of down to the mixture temperature of 30.0 °C: (1500 − 150) / 2500 = 0.54.
MCQ 8CalculationPractice
During a determination of the specific heat capacity of a solid, some heat escapes from the calorimeter to the room after the hot solid is dropped in, so the recorded mixture temperature is below its ideal adiabatic value. Compared with the true value, the specific heat capacity calculated from the readings will be:
Show answer and why every option is right or wrong
Answer: B. Option B. In c = (m_w + W)(T_f − T_cold) / [m_s(T_hot − T_f)], a lowered T_f shrinks the numerator and enlarges the denominator at the same time, so the result is pushed down — the reason the insulating jacket and the brisk transfer are insisted on in the NCERT Physics Lab Manual Class 11, Part 7, page 119.
Why A is wrong: A is wrong because the escaped heat never reaches the water or the calorimeter at all; it is subtracted from the gain side only, so the two sides are not affected equally.
Why C is wrong: C is wrong because an over-estimate needs extra heat arriving on the gain side — hot water droplets carried over with the solid, for example — not heat leaking away.
Why D is wrong: D is wrong because the direction is fixed by the algebra: every unit of heat lost to the room lowers T_f, and T_f appears in the numerator and denominator with opposite effect, so the bias is one-way.
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How do you solve a Experiment 09 Specific Heat Capacity question? A worked example
- 1
Given.
Mass of solid m_s = 1.00 × 10² g, taken from a steam chamber at T_hot = 100.0 °C. Mass of water in the calorimeter m_w = 1.50 × 10² g. Water equivalent of calorimeter and stirrer W = 1.5 × 10¹ g. Initial temperature of water and calorimeter T_cold = 22.0 °C. Mixture temperature T_f = 30.0 °C. Specific heat capacity of water = 1 cal g⁻¹ °C⁻¹ (exact, by definition of the calorie).
- 2
Required.
The specific heat capacity of the solid, in cal g⁻¹ °C⁻¹ and in J kg⁻¹ K⁻¹.
- 3
Concept.
Method of mixtures under an assumed adiabatic enclosure: the heat given out by the solid as it cools to the mixture temperature is taken in by the water and by the calorimeter-plus-stirrer as they warm to that same temperature. The calorimeter enters through its water equivalent, which simply adds to the water mass.
- 4
Formula.
m_s c_s (T_hot − T_f) = (m_w + W) × 1 × (T_f − T_cold).
- 5
Substitution.
100 × c_s × (100.0 − 30.0) = (150 + 15) × (30.0 − 22.0).
- 6
Calculation.
Right side = 165 × 8.0 = 1.32 × 10³ cal. Left side = 100 × c_s × 70.0 = 7.00 × 10³ × c_s. Hence c_s = 1320 / 7000 = 0.1886 cal g⁻¹ °C⁻¹. The specific heat capacity of water (1 cal g⁻¹ °C⁻¹) and the conversion 1 cal = 4.2 J are exact defined values here and do not limit the significant figures; the measured masses and temperatures, given to two and three significant figures, do.
- 7
Final answer.
c_s = 0.19 cal g⁻¹ °C⁻¹, or 0.19 × 4.2 × 10³ = 7.9 × 10² J kg⁻¹ K⁻¹ (two significant figures).
- 8
Common trap.
Dropping the water equivalent gives 150 × 8.0 / 7000 = 0.17 cal g⁻¹ °C⁻¹, about 9 % low. The error is systematic and always downward, because the missing term sits on the gain side of the equation. The same sign of error comes from a slow transfer, which lets the solid cool in the air before it reaches the water.
- 9
Similar NEET-style question.
A calorimeter of water equivalent 1.0 × 10¹ g holds 2.00 × 10² g of water at 25.0 °C. A metal block of mass 8.0 × 10¹ g at 100.0 °C is dropped in and the mixture settles at 29.0 °C. Find the specific heat capacity of the metal in J kg⁻¹ K⁻¹. (Answer: 840 cal / (80 × 71.0) = 0.148 cal g⁻¹ °C⁻¹ ≈ 6.2 × 10² J kg⁻¹ K⁻¹.)
What to remember before solving Experiment 09 Specific Heat Capacity questions
More in Experimental Skills: 6 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
Experiment 09 Specific Heat Capacity questions from past NEET papers
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Sources
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