Can a system be heated and its temperature remains constant?

NCERT Exemplar · Q12

Q. Can a system be heated and its temperature remains constant?

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Concept understanding — First Law of Thermodynamics

First Law of Thermodynamics

The Intuition: Energy is a Bank Account

Imagine you have a bank account. You can deposit money into it, withdraw money from it, or leave it untouched. The total amount of money in your account changes only when money goes in or comes out. You cannot create money from nothing, and money does not vanish into thin air.

Energy works exactly the same way. In any physical or chemical process, energy is never created and never destroyed. It only moves from one place to another, or changes from one form into another. This is the deepest idea behind the First Law.

Now, in thermodynamics, we focus on a specific "bank account": the internal energy of a system. Internal energy ( $U$ ) is the total energy stored inside a substance — the kinetic energy of its molecules jiggling around, plus the potential energy stored in the bonds between them.

If you want to change how much energy is stored inside a system, you have exactly two ways to do it:

Heat ( $Q$ ) — energy that flows because of a temperature difference. Like putting a cold pan on a hot stove.
Work ( $W$ ) — energy transferred by a force moving something. Like pushing a piston to compress a gas.

That's it. No third option. Every change in internal energy comes from either heat or work.

The Precise Statement

\Delta U = Q - W

Where:

$\Delta U$ = change in internal energy of the system
$Q$ = heat added to the system (positive if heat flows in)
$W$ = work done by the system (positive if the system does work on surroundings)

This sign convention is the standard one used in Indian exams (JEE, NEET, etc.). Heat added to the system is positive. Work done by the system is positive.

Watch out

Some textbooks use $Q = \Delta U + W$ or $\Delta U = Q + W$ with a different sign for work. Always check which convention your exam follows. The one above ( $\Delta U = Q - W$ ) is the most common in Indian syllabi.

What This Equation Really Says

Think of it as a balance sheet:

If you add heat ( $Q > 0$ ), internal energy tends to increase.
If the system does work ( $W > 0$ ), internal energy tends to decrease (because energy leaves the system to do the work).
The net change is simply: what came in minus what went out.

If $\Delta U = 0$ , the system has returned to its original internal energy — but that does not mean nothing happened. Heat could have come in, and exactly the same amount of energy could have left as work. The energy just passed through.

Note

| Process | $Q$ | $W$ | $\Delta U$ |

|---------|-----|-----|------------|

| Gas heated at constant volume | + | 0 | Positive |

| Gas cooled at constant volume | – | 0 | Negative |

A Concrete Example

Take a gas trapped in a cylinder with a movable piston. You place the cylinder on a hot plate.

Heat $Q = +100\ \text{J}$ flows into the gas.
The gas expands, pushing the piston upward, doing work $W = +40\ \text{J}$ on the surroundings.

What happens to the internal energy?

$\Delta U = 100 - 40 = +60\ \text{J}$ …

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Yes — during an isothermal process, heat added to a system does work on the surroundings instead of raising the internal energy, so temperature stays constant. The First Law of Thermodynamics explains why.

The question touches a common confusion: we instinctively think "heat in = temperature up." But thermodynamics is more subtle. The First Law says:

\Delta U = Q - W

where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system, and $W$ is work done by the system. Temperature is directly linked to internal energy — for an ideal gas, $U \propto T$ . So if $T$ is constant, $\Delta U = 0$ , and the First Law becomes:

0 = Q - W \quad \Rightarrow \quad Q = W

Every joule of heat added is exactly balanced by work done by the system. No energy is left over to raise the temperature.

The key condition: isothermal process.

"Isothermal" means constant temperature. For this to happen, the system must be in thermal contact with a large reservoir (like a water bath) that absorbs or supplies heat to keep $T$ fixed. The process must also be slow enough that the system stays in equilibrium.
What happens to the heat?

When you add heat $Q$ to an ideal gas in an isothermal expansion, the gas expands against an external pressure. That expansion does work $W = \int P\,dV$ . Since $\Delta U = 0$ , all the heat is converted into work. The gas does not get hotter — it just pushes the piston outward.
A concrete example: isothermal expansion of an ideal gas.

For $n$ moles of an ideal gas at temperature $T$ , the work done in a reversible isothermal expansion from volume $V_i$ to $V_f$ is:

W = nRT \ln\left(\frac{V_f}{V_i}\right)

The heat absorbed is exactly the same:

Q = nRT \ln\left(\frac{V_f}{V_i}\right)

Temperature $T$ appears in the formula, but it does not change during the process.

Watch out

A common mistake is to think "heat always raises temperature." That is false — heat can be entirely converted into work if the internal energy stays constant. The First Law is the boss: $\Delta U = Q - W$ , not $\Delta U = Q$ .

What about real gases or phase changes? …

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Q.Can a system be heated and its temperature remains constant?

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