Q.A metal block of mass 0.20 kg at 150 degree C is dropped in a copper calorimeter of water equivalent 0.025 kg and containing 0.15 kg of water at 27 degree C. If the final temperature is 40 degree C, then calculate the specific heat of the metal. OR What do you mean by 'mean free path' of molecules in a gas? You know that it is expressed as l = 1 / (sqrt(2) * n * pi * d^2). What are n and d here? Draw a neat diagram to show the volume swept by a molecule in time delta_t in which any molecule will collide with it.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →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=mΔTQ
Where:
- c = specific heat capacity (J/kg·°C or J/kg·K)
- Q = heat energy supplied (J)
- m = mass of the substance (kg)
- ΔT = 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 J/kg⋅°C (or ≈ 4200 J/kg·°C in many problems)
- Iron: c≈450 J/kg⋅°C
- Copper: c≈390 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 ΔT.
A Common Mistake to Avoid
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 cwater=4200 J/kg⋅°C)
Solution:
- m=2 kg
- ΔT=50−20=30 °C
- c=4200 J/kg⋅°C
Q=mcΔT=2×4200×30=252000 J=252 kJ
Answer: 252 kJ of heat is required.
Why This Matters …
Using calorimetry (heat lost by the hot metal = heat gained by water + calorimeter), the metal's specific heat can be calculated. In the OR alternative, mean free path relates a gas's molecular density and molecular size to the average distance between collisions. …
Main: specific heat of the metal ≈433 J kg−1K−1. OR: in l=2nπd21, n is molecule number-density and d is molecular diameter.
Main part. By the principle of calorimetry (assuming no heat loss to surroundings): Heat lost by the hot metal block = Heat gained by (water + calorimeter, represented by its water equivalent). mmetals(Tmetal−Tf)=(mwater+w)swater(Tf−Ti), where swater=4200 J kg−1K−1. Substituting: 0.20×s×(150−40)=(0.15+0.025)×4200×(40−27). LHS: 0.20×s×110=22s. RHS: 0.175×4200×13=735×13=9555. So 22s=9555 ⇒ s=229555≈434.3 J kg−1K−1 (close to the specific heat of iron, ≈450 J kg−1K−1 — a physically sensible result for 'a metal').
…
Showing the 12 most recent of 13 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.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 …
- CBSE 2026Set ANNUAL1 markQ.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 ha …
- CBSE 2026Set ANNUAL1 markMCQQ.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).
…
- CBSE 2025Set ANNUAL1 markMCQQ.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×cwater=S⇒W=cwaterS
…
- CBSE 2025Set ANNUAL1 markMCQQ.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:
Heat lost by hot body=Heat gained by cold body
…
- CBSE 2025Set ANNUAL1 markMCQQ.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.
…
- CBSE 2025Set ANNUAL1 markQ.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 …
- CBSE 2025Set ANN1 markQ.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).
…
- CBSE 2024Set ANNUAL1 markMCQQ.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).
…
- CBSE 2024Set SET-AP55001 markQ.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)
…
- CBSE 2024Set ANNUAL1 markMCQQ.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).
…
- CBSE 2023Set ANNUAL1 markMCQQ.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).
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.