Q.Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35 °C to 55 °C. Molar heat capacity of Al is 24 J mol$^{-1}$ K$^{-1}$.

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 …