Q.A geyser heats water flowing at the rate of $3.0$ litres per minute from $27\,^{\circ}\text{C}$ to $77\,^{\circ}\text{C}$. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 \times 10^{4}\ \text{J/g}$?

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

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

1. 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.

2. 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.

3. 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

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:

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

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

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.