A $10\ \text{kW}$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0\ \text{kg}$. How much is the rise in temperature of the block in $2.5$ minutes, assuming $50\%$ of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium $= 0.91\ \text{J g}^{-1}\ \text{K}^{-1}$.

A 10kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0kg. How much is the rise

Exercises · 10.12

10\ \text{kW}

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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 …

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The key idea is to convert the useful work done by the drilling machine into heat absorbed by the aluminium block, then use the specific heat capacity to find the temperature rise. The final temperature rise is $103\ ^\circ\text{C}$ (or $103\ \text{K}$ ).

Concept & Intuition

When a drilling machine bores into a block, the mechanical work done against friction and cutting forces is almost entirely converted into heat. Not all the electrical power drawn by the machine goes into heating the block — some is lost to the machine itself and the surroundings. Here, only 50% of the $10\ \text{kW}$ power actually heats the aluminium block.

The specific heat capacity $c$ tells us how much heat energy is needed to raise the temperature of 1 gram of a substance by 1 K (or $1^\circ\text{C}$ ). For aluminium, $c = 0.91\ \text{J g}^{-1}\text{K}^{-1}$ . Notice the units: grams, not kilograms. That’s a common trap — we must convert the mass of the block from kg to g before using this $c$ .

The relationship is:

Q = m c \Delta T

where $Q$ is the heat absorbed, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the rise in temperature.

We know the useful power (the part that heats the block) and the time for which it operates. So we can find $Q$ , then solve for $\Delta T$ .

Step-by-step solution

1. Find the total electrical energy consumed by the machine in 2.5 minutes.

Power $P = 10\ \text{kW} = 10 \times 10^3\ \text{W} = 10^4\ \text{W}$ .

Time $t = 2.5\ \text{minutes} = 2.5 \times 60\ \text{s} = 150\ \text{s}$ .

Total energy consumed:

E_{\text{total}} = P \times t = 10^4\ \text{W} \times 150\ \text{s} = 1.5 \times 10^6\ \text{J}

2. Determine the energy that actually heats the aluminium block.

Only 50% of the power is used to heat the block. So the useful heat $Q$ is:

Q = 50\% \text{ of } E_{\text{total}} = \frac{50}{100} \times 1.5 \times 10^6\ \text{J} = 0.5 \times 1.5 \times 10^6\ \text{J} = 7.5 \times 10^5\ \text{J}

Watch out

A common mistake is to forget that the specific heat is given in $\text{J g}^{-1}\text{K}^{-1}$ , not $\text{J kg}^{-1}\text{K}^{-1}$ . If you use the mass in kilograms directly, you’ll get an answer that is 1000 times too small. Always check units before plugging numbers. …

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