Which will require more energy: heating a 2.0 kg block of lead by 30 K or heating a 4.0 kg block of copper by 5 K? (s(lead) = 128 J kg⁻¹ K⁻¹, s(copper) = 387 J kg⁻¹ K⁻¹)

Question

Accepted Answer

Concept understanding — Specific Heat Capacity

What is Specific Heat Capacity?

Imagine you have two identical stoves, two identical pots, and you put 1 kg of water in one pot and 1 kg of iron in the other. You turn both stoves to the same flame. After 2 minutes, the iron is scorching hot — you can't touch it. The water is still lukewarm.

Why? Because different substances need different amounts of heat to raise their temperature by the same amount. That's the core idea behind specific heat capacity.

The Intuition

Think of heat as "energy currency" and temperature rise as "buying a degree." Some materials are "cheap" — a little heat buys a big temperature rise. Others are "expensive" — you need to spend a lot of heat to get even a small rise.

Iron is cheap: a small heat input → large temperature jump.
Water is expensive: a large heat input → small temperature jump.

This "expensiveness" is what we call specific heat capacity. It tells you how much heat energy is needed to raise the temperature of 1 kg of a substance by 1 °C (or 1 K).

The Precise Definition

c = \frac{Q}{m \Delta T}

Where:

$c$ = specific heat capacity (J/kg·°C or J/kg·K)
$Q$ = heat energy supplied (J)
$m$ = mass of the substance (kg)
$\Delta T$ = change in temperature (°C or K)

In words: Specific heat capacity is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin).

Key Points to Remember

It's a property of the material, not the object. A small iron nail and a giant iron beam have the same $c$ value — but the beam needs more total heat because it has more mass.
Units matter. Common values:
- Water: $c = 4186 \text{ J/kg·°C}$ (or ≈ 4200 J/kg·°C in many problems)
- Iron: $c \approx 450 \text{ J/kg·°C}$
- Copper: $c \approx 390 \text{ J/kg·°C}$
Notice water's value is about 10 times that of iron — that's why water heats up so slowly compared to metals.
The formula works both ways. If a substance cools down, it releases the same amount of heat it would absorb to warm up by the same $\Delta T$ .

A Common Mistake to Avoid

Watch out

Don't confuse specific heat capacity ( $c$ ) with heat capacity ( $C$ ). Heat capacity is for an entire object: $C = mc$ . A large iron block can have a higher heat capacity than a small cup of water, even though iron's $c$ is much smaller. Always check: are we talking about per kg or for the whole thing?

Worked Example (Exam-Style)

Problem: How much heat is needed to raise the temperature of 2 kg of water from 20 °C to 50 °C? (Take $c_{\text{water}} = 4200 \text{ J/kg·°C}$ )

Solution:

$m = 2 \text{ kg}$
$\Delta T = 50 - 20 = 30 \text{ °C}$
$c = 4200 \text{ J/kg·°C}$

Q = mc\Delta T = 2 \times 4200 \times 30 = 252\,000 \text{ J} = 252 \text{ kJ}

Answer: 252 kJ of heat is required.

Why This Matters

Specific heat capacity explains countless everyday phenomena:

Why coastal cities have milder climates than inland ones (water's high $c$ stores heat)
Why a metal spoon in hot tea gets hot instantly while the tea stays hot longer
Why car radiators use water as coolant — it absorbs lots of heat without boiling

Important

Water has one of the highest specific heat capacities of any common substance. This single fact explains why oceans regulate Earth's climate, why your body uses water to maintain temperature, and why water is the go-to coolant in engines and power plants.

Final takeaway: Specific heat capacity is the "thermal inertia" of a material — how stubbornly it resists changing temperature when you add or remove heat. The higher the $c$ , the more heat you need to move its temperature.

A quick web search for "Specific Heat Capacity class 12 physics" or "Specific Heat Capacity class 11 physics" turns up this exact idea, because Specific Heat Capacity sits squarely within the Thermal Properties of Matter coverage of NCERT Class 11 Physics, so it is fair game for both CBSE board questions and competitive-exam numericals. Revisiting the NCERT Physics textbook exercises for this chapter alongside the walkthrough above is a solid way to convert this into exam-ready practice.

Using the heat equation $Q=m\,\Delta T\,s$ (Eq. 7.28) for each block: for lead, $Q_{lead} = 2\text{ kg}\times30\text{ K}\times128\text{ J kg}^{-1}\text{K}^{-1} = 7680\text{ J}$ ; for copper, $Q_{copper} = 4\text{ kg}\times5\text{ K}\times387\text{ J kg}^{-1}\text{K}^{-1} = 7740\text{ J}$ . Since $7740\text{ J} > 7680\text{ J}$ , heating the copper block requires slightly MORE energy, even though its temperature rise (5 K) is much smaller than lead's (30 K) -- because copper's specific heat capacity is roughly three times larger and its mass is double, more than compensating for the smaller temperature rise. [!ANSWER] Copper (7740 J vs lead's 7680 J)

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the S.I. unit of specific heat capacity?$
(a) J kg^-2 K^-1
(b) J kg^-1 K^-1
(c) J^-1 kg^-1 K^-1
(d) J kg^-1 K^-2
›Reveal solutionSolution

The SI unit of specific heat capacity is J kg⁻¹ K⁻¹.

Specific heat capacity is defined as the heat required to raise the temperature of unit mass of a substance by one unit (kelvin/°C): s = Q/(mΔT). Since Q is measured in joules (J), m in kilograms (kg), and ΔT in kelvin (K), the unit of s works out to J/(kg·K) = J kg⁻¹ K⁻¹.

✓Final answer
$The correct option is (b) J kg⁻¹ K⁻¹.$
CBSE 2026Set ANNUAL1 mark
$Q. Write the definition of Specific Heat.$
›Reveal solutionSolution

Specific heat s = Q/(mΔT) — the heat needed per unit mass per unit rise in temperature.

When an amount of heat Q is supplied to a body of mass m, its temperature rises by some amount ΔT (as long as no phase change occurs). The specific heat (or specific heat capacity) of the substance is defined as the heat required to raise the temperature of a unit mass of that substance by one unit (one kelvin or one degree Celsius): s = Q/(mΔT), with SI unit J kg⁻¹ K⁻¹. It is an intrinsic property of the material — e.g. water has an unusually high specific heat (≈4186 J kg⁻¹ K⁻¹) compared to most metals, which is why water heats up and cools down slowly.

✓Final answer
$Specific heat s = Q/(mΔT) — heat required per unit mass to raise temperature by 1 K.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. SI unit of specific heat capacity is$
(a) J kg K^-1
(b) J kg^-1 K^-1
(c) J kg^-1 K^-2
(d) J mol^-1 K^-1
›Reveal solutionSolution

SI unit of specific heat capacity is J kg^-1 K^-1. Answer (B).

Specific heat capacity c is defined by Q = m c dT, so c = Q/(m dT).

Units: joule/(kilogram x kelvin) = J kg^-1 K^-1.

✓Final answer
$(B) J kg^-1 K^-1.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The heat capacity of a vessel is 520 cal gm^-1. Its water equivalent will be (A) 520 cal (B) 520 gm (C) 52 gm (D) none of these$
›Reveal solutionSolution

A vessel with heat capacity 520 cal/°C has a water equivalent of 520 g.

Heat capacity (thermal capacity) of a body is the amount of heat required to raise its temperature by 1°C: $S = mc$ (units cal/°C).

Water equivalent $W$ is defined as the mass of water that would absorb (or need) the same amount of heat as the given body for the same temperature change:

$W \times c_{water} = S \Rightarrow W = \dfrac{S}{c_{water}}$

Since the specific heat of water $c_{water} = 1\text{ cal g}^{-1}\,{}^\circ\text{C}^{-1}$ , the water equivalent (in grams) is numerically equal to the heat capacity (in cal/°C):

$W = \dfrac{520\text{ cal/°C}}{1\text{ cal g}^{-1}\,{}^\circ\text{C}^{-1}} = 520\text{ g}$

✓Final answer
$(B) 520 gm.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. On which law is the principle of calorimetry based? (A) The law of energy conservation (B) The law of momentum conservation (C) The law of angular momentum conservation (D) None of these$
›Reveal solutionSolution

The principle of calorimetry (heat lost = heat gained) is a direct application of the law of conservation of energy.

When a hot and a cold substance are mixed in an isolated system (no heat exchange with surroundings), the principle of calorimetry states:

$\text{Heat lost by hot body} = \text{Heat gained by cold body}$

This is simply energy conservation applied to a closed, isolated thermal system — the total thermal energy of the system remains constant, it is merely redistributed between the two substances until they reach a common equilibrium temperature.

✓Final answer
$(A) The law of energy conservation.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. If Cp and Cv are the specific heat capacity at constant pressure and specific heat capacity at constant volume, then which option is correct?$
(a) Cp < Cv
(b) Cp > Cv
(c) Cp = Cv
(d) Cp ~ Cv
›Reveal solutionSolution

Cp (specific heat at constant pressure) is always greater than Cv (specific heat at constant volume) because at constant pressure some of the supplied heat goes into doing expansion work, not just raising temperature.

At constant volume, all the heat supplied to a gas goes entirely into increasing its internal energy (raising temperature), since the gas cannot expand and do work.

At constant pressure, when heat is supplied, the gas both raises its internal energy AND expands, doing work against the constant external pressure. So more heat is required to raise the temperature by the same amount.

Hence Cp > Cv always (for an ideal gas, Cp - Cv = R, Mayer's relation, which is positive).

✓Final answer
$(b) Cp > Cv.$
CBSE 2025Set ANNUAL1 mark
$Q. Answer in one word or one sentence: Write the value of specific heat of water in SI unit.$
›Reveal solutionSolution

Water's specific heat capacity is about 4186 J kg^-1 K^-1 in SI units.

Specific heat capacity (c) is the amount of heat required to raise the temperature of a unit mass of a substance by one degree (kelvin or Celsius). For water, this value is unusually high compared to most substances, which is why water is an effective coolant and moderates climate near large water bodies. In SI units, c(water) is approximately 4186 J kg^-1 K^-1 (sometimes rounded to 4200 J kg^-1 K^-1, or historically quoted as 1 cal g^-1 degreeC^-1).

✓Final answer
$The specific heat of water in SI unit is approximately 4186 J kg^-1 K^-1.$
CBSE 2025Set ANN1 mark
$Q. Name the substance which has the highest specific heat capacity.$
›Reveal solutionSolution

Water has the highest specific heat capacity among common substances, about 4186 J per kg per kelvin.

Specific heat capacity c is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1 degree C): Q = m c (delta T).

Water is remarkable because it needs a very large amount of heat for even a small temperature rise - about 4186 J kg^-1 K^-1, far higher than most solids and liquids. This large value is why water is used as a coolant and why coastal climates are moderate.

✓Final answer
$Water (about 4186 J kg^-1 K^-1).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following materials has the highest specific heat capacity?$
(a) Ice
(b) Glass
(c) Iron
(d) Water
›Reveal solutionSolution

Water's specific heat capacity (~4186 J/kg·K, i.e. 1 cal/g·°C) is exceptionally high compared to other common substances — much higher than ice, glass, or iron.

Approximate specific heat capacities: Water ≈ 4186 J/(kg·K); Ice ≈ 2100 J/(kg·K); Glass ≈ 840 J/(kg·K); Iron ≈ 450 J/(kg·K).

Water's high specific heat is why large water bodies moderate coastal climates and why water is used as a coolant — it can absorb/release large amounts of heat with only a small temperature change.

✓Final answer
$(d) Water.$
CBSE 2024Set SET-AP55001 mark
$Q. What is specific heat capacity?$
›Reveal solutionSolution

Specific heat capacity (c) is the heat required to raise the temperature of 1 kg (unit mass) of a substance by 1°C (or 1 K).

When an amount of heat Q is supplied to a body of mass m and its temperature rises by ΔT, the heat capacity of the whole body is Q/ΔT (heat needed per degree). Dividing further by the mass gives a property that depends only on the MATERIAL, not on how much of it there is:

c = Q / (m ΔT)

This is the specific heat capacity. Its SI unit is J kg^-1 K^-1. Different substances have different specific heat capacities (e.g., water has an unusually high one, about 4186 J kg^-1 K^-1), which is why water heats up and cools down slowly compared to most other materials.

✓Final answer
$Specific heat capacity c = Q/(mΔT) — the heat needed to raise the temperature of unit mass of a substance by one degree.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The SI unit for specific heat capacity is:$
(a) J kg^-1 K^-1
(b) J kg^-1
(c) K kg^-1 J^-1
(d) J kg K^-1
›Reveal solutionSolution

Specific heat capacity s is defined by Q = m s ΔT, so s = Q/(m ΔT), giving SI unit J kg^-1 K^-1.

Specific heat capacity is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one unit (one kelvin).

From Q = m s ΔT, we get s = Q / (m ΔT).

SI unit of Q is joule (J), of m is kilogram (kg), and of ΔT is kelvin (K).

So the SI unit of s = J / (kg × K) = J kg^-1 K^-1.

✓Final answer
$The SI unit of specific heat capacity is J kg^-1 K^-1 — option (a).$
CBSE 2023Set ANNUAL1 markMCQ
$Q. SI unit of specific heat capacity is:$
(a) cal/g/k
(b) kcal/g
(c) J/kg/k
(d) J/kg
›Reveal solutionSolution

SI unit of specific heat capacity = J kg^-1 K^-1 (J/kg/K).

Specific heat capacity c is defined by Q = mcΔT, so c = Q/(mΔT).

Units: joule/(kilogram × kelvin) = J kg^-1 K^-1, written J/kg/K.

✓Final answer
$(C) J/kg/k.$

Q.Which will require more energy: heating a 2.0 kg block of lead by 30 K or heating a 4.0 kg block of copper by 5 K? (s(lead) = 128 J kg⁻¹ K⁻¹, s(copper) = 387 J kg⁻¹ K⁻¹)

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