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Specific heat of solid and liquid by method of mixtures: NEET MCQs with answers

50 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In a laboratory experiment using the method of mixtures, a calorimeter contains a liquid. If the mass of the calorimeter is m, its specific heat capacity is c, and its water equivalent is W, which of the following expressions correctly relates W to m and c? (Take the specific heat capacity of water as c_w)

  1. W = (m · c) / c_w
  2. W = m · c · c_w
  3. W = (m · c_w) / c
  4. W = m / (c · c_w)
Show answer and explanation

Answer: (A) W = (m · c) / c_w

Water equivalent W is defined as the mass of water that absorbs or releases the exact same amount of heat as the substance for a unit rise or fall in temperature. Hence, W · c_w · ΔT = m · c · ΔT, which gives W = (m · c) / c_w.

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2. A hot solid of mass m₁ and specific heat s₁ at T₁ is dropped into water (m₂, s₂) at T₂ in a calorimeter (mc, sc). At final temperature T:

  1. m₁s₁T₁ = m₂s₂T₂
  2. m₁s₁(T - T₁) = m₂s₂(T₂ - T)
  3. m₁(T₁ - T) = m₂(T - T₂)
  4. m₁s₁(T₁ - T) = (m₂s₂ + mcsc)(T - T₂)
Show answer and explanation

Answer: (D) m₁s₁(T₁ - T) = (m₂s₂ + mcsc)(T - T₂)

Heat lost by the solid equals heat gained by the water and the calorimeter, which gives this equation for the unknown specific heat.

NCERT Class 11

hard

3. In a Regnault's apparatus experiment, the temperature of 100 g of hot metal pellets is measured to be 100 °C. While quickly transferring them into 100 g of water at 20 °C, 0.5 g of boiling water (at 100 °C) is accidentally carried over with the pellets into the calorimeter. Taking specific heat of water as 4.2 J/(g °C), the excess heat carried into the mixture solely due to this carried-over water relative to a final temperature of 25 °C is:

  1. 42 J
  2. 84 J
  3. 315 J
  4. 157.5 J
Show answer and explanation

Answer: (D) 157.5 J

The 0.5 g of carried-over water cools from 100 °C to the equilibrium temperature of 25 °C. The heat released by this water is Q = m × c_w × ΔT = 0.5 g × 4.2 J/(g °C) × (100 - 25) °C = 2.1 × 75 = 157.5 J. This causes an overestimation of the heat lost by the solid.

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4. In a method of mixtures experiment, why is it recommended to use a stirrer made of the exact same material as the calorimeter?

  1. To avoid thermo-electric currents generated by dissimilar metals
  2. To accelerate convective currents inside the calorimeter
  3. To ensure the stirrer has a much higher specific heat capacity than the liquid
  4. Because its mass can be directly added to the calorimeter's mass when calculating thermal capacity
Show answer and explanation

Answer: (D) Because its mass can be directly added to the calorimeter's mass when calculating thermal capacity

When the stirrer is made of the same material as the calorimeter vessel (typically copper), their masses can simply be added together (m_total = m_calorimeter + m_stirrer) and multiplied by a single specific heat capacity, simplifying calculations and reducing source of error.

easy

5. Which of the following modifications best minimizes error due to heat loss by convection in a calorimetry experiment?

  1. Using a highly polished outer surface for the calorimeter
  2. Increasing the volume of water used inside the calorimeter
  3. Enclosing the calorimeter inside a double-walled chamber with an insulating lid
  4. Using a stirrer with very large surface area
Show answer and explanation

Answer: (C) Enclosing the calorimeter inside a double-walled chamber with an insulating lid

A double-walled enclosure with an insulating lid creates a stagnant air layer or vacuum around the calorimeter and prevents air currents, thereby minimizing heat loss through convection. Polishing surfaces reduces radiation, not convection.

hard

6. In a calorimeter experiment, the recorded equilibrium temperature of the mixture is T_m, the initial temperature of water is T_w, and the room temperature is T_r. To apply Rumford's cooling correction effectively, how should T_w be ideally chosen relative to room temperature?

  1. T_w should be equal to T_r
  2. T_w should be as much below T_r as T_m is above T_r
  3. T_w should be much higher than T_r
  4. T_w should be 0 °C regardless of room temperature
Show answer and explanation

Answer: (B) T_w should be as much below T_r as T_m is above T_r

In Rumford's method, the initial temperature of the liquid is kept below room temperature by approximately the same amount that the final temperature rises above room temperature. This ensures heat gained from the surroundings in the first half approximately equals heat lost to surroundings in the second half, cancelling the radiation error.

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7. A student carries out the method of mixtures to find the specific heat of a metal. If some hot water drops cling to the metal while transferring it from the boiling water bath into the calorimeter, the calculated value of the specific heat of the metal will be:

  1. Equal to the actual value
  2. Smaller than the actual value
  3. Greater than the actual value
  4. Zero
Show answer and explanation

Answer: (C) Greater than the actual value

The carried-over hot water releases additional heat (due to the very high specific heat of water: 4.2 J g⁻¹ °C⁻¹) into the calorimeter, resulting in a higher observed final temperature θ. In the formula s = Heat gained / [m(T - θ)], the numerator is overestimated while the denominator is slightly decreased, leading to an overestimation of the specific heat.

easy

8. An alloy sphere of mass m and specific heat s is heated to temperature T₁ and immersed in mass m of water at temperature T₂ in a calorimeter of negligible heat capacity. If the specific heat of the alloy is half that of water, the equilibrium temperature T_e is:

  1. (T₁ + 2T₂) / 3
  2. (2T₁ + T₂) / 3
  3. (T₁ + T₂) / 2
  4. (T₁ + 3T₂) / 4
Show answer and explanation

Answer: (A) (T₁ + 2T₂) / 3

Heat lost by alloy = m × s × (T₁ - T_e). Heat gained by water = m × s_w × (T_e - T₂). Given s = s_w / 2, we have: m × (s_w / 2) × (T₁ - T_e) = m × s_w × (T_e - T₂). Dividing by m × s_w: (T₁ - T_e) / 2 = T_e - T₂ => T₁ - T_e = 2 T_e - 2 T₂ => 3 T_e = T₁ + 2 T₂ => T_e = (T₁ + 2T₂) / 3.

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9. During an experiment to determine the specific heat of a metal using the method of mixtures, a student vigorously stirs the water in the calorimeter with a mechanical stirrer for a prolonged time after dropping the hot metal. How does this action affect the calculated value of the specific heat capacity of the metal?

  1. It results in an underestimation because vigorous stirring cools the liquid via evaporation.
  2. It results in an underestimation of the specific heat capacity because vigorous stirring enhances convective heat loss.
  3. It does not affect the result because the mechanical energy supplied is entirely negligible.
  4. It results in an overestimation of the specific heat capacity because mechanical work adds extra heat to the liquid.
Show answer and explanation

Answer: (D) It results in an overestimation of the specific heat capacity because mechanical work adds extra heat to the liquid.

Vigorous and prolonged mechanical stirring does significant work against viscous forces in the liquid, converting mechanical energy into thermal energy. This artificial rise in temperature leads to an overestimated final equilibrium temperature, making the apparent heat gained by the water greater, which directly causes an overestimation of the metal's specific heat capacity.

easy

10. A student determines the specific heat capacity of a solid metal cylinder using the method of mixtures. The formula used is s = [(m_c · s_c + m_w · s_w)(T_f - T_i)] / [m_s(T_s - T_f)]. If the student erroneously records a value of the initial temperature of the hot solid (T_s) that is higher than its true value, how does this affect the calculated value of s?

  1. The calculated s will be smaller than the true value.
  2. The calculated s will be greater than the true value.
  3. The calculated s will remain unchanged.
  4. The calculated s may increase or decrease depending on the mass of water.
Show answer and explanation

Answer: (A) The calculated s will be smaller than the true value.

From the formula, the specific heat s is inversely proportional to (T_s - T_f). If the recorded T_s is greater than its true value, the denominator (T_s - T_f) becomes artificially large, which causes the calculated value of s to be smaller than the true value.

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11. A piece of iron of mass 200 g at 100 °C is dropped into 200 g of an unknown liquid at 15 °C contained in a vessel whose thermal capacity is negligible. The mixture settles at 25 °C. Given specific heat of iron is 0.11 cal g⁻¹ °C⁻¹, calculate the specific heat capacity of the liquid.

  1. 0.275 cal g⁻¹ °C⁻¹
  2. 0.550 cal g⁻¹ °C⁻¹
  3. 0.412 cal g⁻¹ °C⁻¹
  4. 0.825 cal g⁻¹ °C⁻¹
Show answer and explanation

Answer: (D) 0.825 cal g⁻¹ °C⁻¹

Heat lost by iron = 200 × 0.11 × (100 - 25) = 22 × 75 = 1650 cal. Heat gained by liquid = 200 × s_liq × (25 - 15) = 200 × s_liq × 10 = 2000 s_liq. Equating: 2000 s_liq = 1650, so s_liq = 1650 / 2000 = 0.825 cal g⁻¹ °C⁻¹.

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12. During the experiment the mixture is stirred continuously in order to:

  1. Increase the mass
  2. Maintain a uniform temperature throughout
  3. Increase heat loss
  4. Cool the mixture faster
Show answer and explanation

Answer: (B) Maintain a uniform temperature throughout

Stirring ensures uniform temperature in the mixture so the measured final temperature is accurate.

NCERT Class 11

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