Temperature and heat

Press a hand to a mug of tea and something crosses from the mug into your skin — you call it "hot." Press the same hand to a steel railing on a winter morning and something seems to leave your skin just as fast — you call that "cold." In both cases what is actually happening is the same physical event running in opposite directions: energy is moving from the object at higher temperature to the object at lower temperature, because they are in contact and not yet at the same temperature. That flowing energy is heat. The property that tells you which way it will flow is temperature. CBSE Class 11 opens the unit Thermal Properties of Matter by insisting you keep these two words apart, because every later idea in the chapter — thermal expansion, specific heat, latent heat, conduction, radiation — is built on the clean separation between "how hot" and "how much energy." The everyday language misleads you constantly. People say a blanket "gives you heat," a thermos "keeps the heat in," a body "has a lot of heat." None of that survives a physics exam. A blanket generates no energy at all; it merely slows the rate at which your body's heat escapes. A body does not "contain" heat — it contains internal energy, and heat is only the name for energy in transit across a boundary because of a temperature difference. Get this distinction sharp and a whole family of numerical problems — mixing two liquids, reading a thermometer, deciding which way heat flows in a machine — become one-line applications of a single balance principle: heat lost by the hotter object equals heat gained by the cooler one, until both reach a common temperature called thermal equilibrium. This article builds that principle from the zeroth law of thermodynamics upward, and works two clean numbers so you can see it computed, not just stated.

Key takeaways

Temperature and heat are different physical quantities. Temperature decides the direction of heat flow; heat is the energy that actually flows because of a temperature difference.
Heat ( $Q$ ) is energy in transit, measured in joules (SI) or calories, with $1\ \text{cal} = 4.186\ \text{J}$ ; it exists only while flowing, never as a stored quantity.
Two bodies are in thermal equilibrium when no net heat flows between them — they are then at the same temperature, and this idea is formalised as the zeroth law of thermodynamics.
The zeroth law makes thermometers possible: if a thermometer (body C) reaches equilibrium with a body A, and separately with body B, then A and B are also in equilibrium with each other, at the same reading.
Kelvin and Celsius readings differ by a fixed offset, $T(\text{K}) = T(^\circ\text{C}) + 273.15$ , but a difference in temperature is numerically identical on both scales — $\Delta T$ in kelvin equals $\Delta T$ in celsius.
Heat lost equals heat gained for two bodies exchanging energy in isolation — the calorimetric balance principle that fixes a common final temperature.
Key idea: temperature is the equilibrium-deciding property that fixes the direction of heat flow, heat is the energy that actually flows because of a temperature difference until a common temperature is reached, and the zeroth law of thermodynamics is what lets a thermometer measure that temperature at all.

The core insight

Temperature tells you which way energy will flow between two bodies in contact; heat is that energy itself, flowing spontaneously from the hotter body to the cooler one until both reach the same temperature — a state called thermal equilibrium, whose logical foundation is the zeroth law of thermodynamics.

Where it fits in the syllabus

This article opens the CBSE Class 11 Physics unit Thermal Properties of Matter, sitting before thermometry, thermal expansion, specific heat and heat transfer — every one of which assumes you already know the difference between heat and temperature. The zeroth law, the joule-calorie relation, and the Celsius-Kelvin conversion given here are core, examinable NCERT content. Detailed thermometer construction, gas-thermometer scales and calibration are deliberately left to the next article in this chapter; here the goal is only to nail down what temperature and heat are before you measure or manipulate them.

The physical picture

Consider two metal blocks placed in contact inside a perfectly insulated box, one initially hotter than the other. Experience — and every calorimetry experiment ever run — shows that energy always moves spontaneously from the hotter block to the cooler one, never the reverse, and that this flow continues only until both blocks reach a single common temperature. Once that common temperature is reached, contact no longer changes either block's temperature, even though both blocks still possess internal energy; what has stopped is the net flow, not the internal energy itself.

Q = mc\,\Delta T

where: $Q$ is heat exchanged (J), $m$ is mass (kg), $c$ is specific heat capacity of the material (J/kg·K), and $\Delta T$ is the temperature change produced by that heat. This relation is the working preview you will formalise fully with $C_p$ and $C_v$ later in the chapter; here it is enough to know that heat exchanged and temperature change are proportional for a body that does not change state, and that this proportionality is what lets you turn "heat flows until equilibrium" into an actual number. The absolute temperature scale used throughout Indian physics classrooms is the kelvin, related to the everyday Celsius scale by a fixed shift:

T(\text{K}) = T(^\circ\text{C}) + 273.15

where: the two scales use degrees of exactly the same size — a rise of one kelvin is a rise of one degree Celsius — so only the zero point differs (0 K is absolute zero, the lowest temperature theoretically attainable, while $0^\circ\text{C}$ is the everyday freezing point of water). This is why a temperature reading must have 273.15 added to convert to kelvin, but a temperature difference never gets that addition — subtracting two kelvin readings that both had 273.15 added just cancels it out again.

Figure

Two insulated metal blocks in contact showing heat flowing from the hotter block at eighty degrees Celsius to the cooler block at twenty degrees Celsius until both reach a common equilibrium temperature of fifty degrees Celsius, with a Celsius to Kelvin scale alongside

Notice the picture is entirely one-directional: nothing in this chapter allows heat to flow spontaneously from cold to hot on its own. That directionality is the qualitative seed of the second law of thermodynamics, whose formal statements you meet properly in the Thermodynamics unit later this year — for now it is enough to treat "heat flows hot to cold, never the reverse, unless external work is done" as an experimental fact you can rely on in every problem.

Derivation AThe common equilibrium temperature from a heat balance

The goal is to find, from first principles, the single common temperature two bodies settle to when they exchange heat only with each other inside a perfectly insulated enclosure, using nothing beyond the proportional relation $Q = mc\,\Delta T$ and the physical fact that an isolated system's total energy exchange with itself must sum to zero — heat lost by one part must exactly equal heat gained by the other, since no energy escapes to the outside and no external work is done on either block.

Step 1. Set up the two bodies. Let body 1 have mass $m_1$ , specific heat $c_1$ , and initial temperature $T_1$ ; body 2 have mass $m_2$ , specific heat $c_2$ , and initial temperature $T_2$ , with $T_1 > T_2$ . Let $T_f$ be the unknown common final temperature they both reach.

Step 2. Write the heat lost by the hotter body. Body 1 cools from $T_1$ down to $T_f$ , so using $Q = mc\,\Delta T$ with the temperature drop as the magnitude, the heat it gives up is $Q_{\text{lost}} = m_1 c_1 (T_1 - T_f)$ . This quantity is positive because $T_1 > T_f$ , consistent with body 1 actually losing energy.

Step 3. Write the heat gained by the cooler body. Body 2 warms from $T_2$ up to $T_f$ , so the heat it absorbs is $Q_{\text{gained}} = m_2 c_2 (T_f - T_2)$ , again positive because $T_f > T_2$ .

Step 4. Apply the isolation condition. Because the enclosure is insulated, no heat escapes to the surroundings and no heat is created or destroyed — every joule that leaves body 1 must be the same joule that enters body 2. This gives the calorimetric balance equation $Q_{\text{lost}} = Q_{\text{gained}}$ , that is, $m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$ .

Step 5. Solve for $T_f$ . Expand both sides: $m_1 c_1 T_1 - m_1 c_1 T_f = m_2 c_2 T_f - m_2 c_2 T_2$ . Collect every $T_f$ term on one side: $m_1 c_1 T_1 + m_2 c_2 T_2 = T_f (m_1 c_1 + m_2 c_2)$ . Dividing through gives the general result

T_f = \frac{m_1 c_1 T_1 + m_2 c_2 T_2}{m_1 c_1 + m_2 c_2}

Step 6. Check the special case that matches everyday intuition. If the two bodies are identical — same mass, same material, so $m_1 c_1 = m_2 c_2$ — the formula collapses to the simple average $T_f = \dfrac{T_1 + T_2}{2}$ , exactly the symmetric halfway point you would expect when neither body has any advantage in "thermal mass." This is the special case used in Worked example 1 below; the general formula is the tool you reach for whenever the two masses or materials differ.

Worked examples

Board level - two identical blocks reaching equilibrium

A metal block at $80^\circ\text{C}$ is placed in contact with an identical metal block at $20^\circ\text{C}$ inside an insulated box, so no heat escapes to the surroundings. Describe what happens to the temperature of each block over time, and state the final equilibrium temperature.

Step 1. Identify the direction of flow. Heat flows from the hotter block ( $80^\circ\text{C}$ ) to the cooler block ( $20^\circ\text{C}$ ) because temperature difference alone drives heat flow — no substance physically moves between them.

Step 2. Track both temperatures. As heat flows, the hot block's temperature falls and the cold block's temperature rises. Since the blocks are identical (same mass, same material, so identical $mc$ ), conservation of energy means the temperature drop of one equals the temperature rise of the other at every instant, so they approach each other symmetrically.

Step 3. Apply the equal-mass special case from the derivation above. With $m_1 c_1 = m_2 c_2$ , the general balance formula reduces to a simple average of the two starting temperatures.

T_f = \frac{80 + 20}{2} = \mathbf{50^\circ C}

They reach this common final temperature where no more net heat flows — this is thermal equilibrium, and the answer matches the "heat lost equals heat gained" balance used to derive it.

Competitive level - unequal masses of the same metal

A $2\ \text{kg}$ block of copper at $90^\circ\text{C}$ is placed in contact with a $5\ \text{kg}$ block of the same copper at $18^\circ\text{C}$ inside an identical insulated enclosure. Both blocks are of the same material, so their specific heats $c$ are equal. Find the common equilibrium temperature, and state which block's temperature changes more.

Step 1. Recognise that identical material lets $c$ cancel out of the general formula, since $c_1 = c_2 = c$ appears in every term of both numerator and denominator.

T_f = \frac{m_1 T_1 + m_2 T_2}{m_1 + m_2}

Step 2. Substitute the given masses and temperatures: $m_1 = 2\ \text{kg}$ , $T_1 = 90^\circ\text{C}$ , $m_2 = 5\ \text{kg}$ , $T_2 = 18^\circ\text{C}$ .

T_f = \frac{(2)(90) + (5)(18)}{2 + 5} = \frac{180 + 90}{7} = \frac{270}{7}

Step 3. Compute the numerical result and round sensibly for a physics answer.

T_f \approx \mathbf{38.6^\circ C}

Step 4. Compare the two temperature changes. The small block cools from $90^\circ\text{C}$ to $38.6^\circ\text{C}$ , a drop of about $51.4^\circ\text{C}$ ; the large block warms from $18^\circ\text{C}$ to $38.6^\circ\text{C}$ , a rise of about $20.6^\circ\text{C}$ . The heavier block changes temperature far less for the same total heat exchanged, because a larger mass needs more heat per degree of change — the equilibrium point sits closer to the temperature of the larger mass, not at the midpoint of the two starting temperatures, since the masses are unequal this time. Timing note: this is a two-line substitution once you know the formula, but the trap is reaching for the equal-mass shortcut $T_f = (T_1+T_2)/2$ out of habit — check whether masses (and specific heats) are actually equal before using it; in a competitive paper this should take under ninety seconds including the sanity check on which block changed more.

Common traps

Saying a body "contains heat" — bodies contain internal energy; heat is energy in transit due to a temperature difference, and once transferred it becomes part of the internal energy of the receiving body, not a separately stored "heat content."
Confusing temperature (intensive — does not depend on the amount of substance) with heat (extensive — depends on mass) — a small cup of boiling water and a large pot of boiling water are at the same temperature, but the pot has absorbed far more heat to reach boiling.
Converting a temperature difference by adding 273.15 — this is wrong; only absolute readings need the offset, since the offset cancels out automatically when you subtract two readings on the same scale.
Reaching for the equal-mass shortcut $T_f = (T_1+T_2)/2$ when the masses or specific heats are actually different — the correct tool is always the weighted balance $T_f = \dfrac{m_1c_1T_1+m_2c_2T_2}{m_1c_1+m_2c_2}$ , and the simple average is only a special case of it.
Thinking heat flow stops only at $0^\circ\text{C}$ or some other fixed value — it stops whenever the two bodies in contact reach the same temperature as each other, whatever numerical value that happens to be.

Common trapHeat is not a substance stored inside a body — it is energy in transit, and only exists while flowing

The single biggest scoring error in this topic is treating heat like an invisible fluid that a hot body "has more of" and a cold body "has less of," something that could in principle be measured sitting inside an object at rest. It cannot: ask "how much heat does this block have right now, at rest, in isolation" and the question is meaningless, because heat is defined only as energy crossing a boundary due to a temperature difference. A body at rest, not exchanging energy with anything, has internal energy (kinetic and potential energy of its molecules) but zero heat, because nothing is flowing. The moment you place it in contact with something at a different temperature, heat begins to flow and only then does the word apply — and it stops applying again the instant equilibrium is reached, even though the internal energy of both bodies is now different from what it was before contact.

You should be able to…

State the difference between heat and temperature in one sentence each, and explain why one is intensive and the other extensive.
Explain thermal equilibrium and state the zeroth law of thermodynamics, including why it justifies using a thermometer.
Convert between Celsius and Kelvin correctly for absolute readings, and recognise that temperature differences need no offset conversion.
Derive and apply the common-equilibrium-temperature formula $T_f = \dfrac{m_1c_1T_1+m_2c_2T_2}{m_1c_1+m_2c_2}$ , including the equal-mass special case.
Predict the direction of spontaneous heat flow between two bodies given only their temperatures.
Identify the SI units of heat (joule) and temperature (kelvin), and convert between joules and calories using $1\ \text{cal} = 4.186\ \text{J}$ .

Ask the tutor: if you are still mixing up heat and temperature, or want to see the equilibrium-temperature formula worked with three different masses or materials, ask the AI tutor to walk through a few more contact-and-balance scenarios with you.

If a hot cup of tea and a large hot pot of tea are at the same temperature, why does the pot burn me worse?

Temperature tells you how hot something is, but it says nothing about how much energy is available to transfer into your skin. The pot has far more mass than the cup, so at the same temperature it holds far more internal energy and can deliver far more heat before it cools down to your skin's temperature. Temperature is an intensive property — it does not scale with the amount of substance — while heat exchanged is extensive and scales directly with mass, through $Q = mc\,\Delta T$ . Equal temperature, very unequal heat delivered: that is exactly the heat-versus-temperature distinction this chapter asks you to master.

Why does heat only flow from hot to cold and never the other way on its own?

This is an experimental fact about how energy spontaneously spreads out, and it is one of the deepest ideas in physics — you will meet its full, formal statement as the second law of thermodynamics later in Class 11. For now, treat it as a rule you can rely on: two bodies at different temperatures placed in contact will always drive net energy from the hotter one to the cooler one until they reach a common temperature, never the reverse, unless something external (like a refrigerator doing work) forces it backwards. Nothing in this chapter asks you to explain why at a molecular level — only to use the rule correctly in balance problems.

Does the choice of Celsius or Kelvin change my answer for Q equals mc delta T?

No, and this is exactly why the distinction between readings and differences matters so much. Because one degree Celsius and one kelvin represent the identical-sized step, a temperature change of, say, $30^\circ\text{C}$ is also a temperature change of $30\ \text{K}$ — no conversion needed. You only need to add or subtract $273.15$ when you are converting an absolute reading (like "the room is at $25^\circ\text{C}$ , what is that in kelvin") — never when you are using a $\Delta T$ inside a formula like $Q = mc\,\Delta T$ .

How is the zeroth law of thermodynamics actually used, not just stated?

Every time you use a thermometer, you are relying on the zeroth law without thinking about it. Put a thermometer (call it body C) in contact with your body (body A) until it stops changing — it has now reached thermal equilibrium with you. Earlier, that same thermometer was calibrated by putting it in equilibrium with reference baths (body B) at known temperatures. The zeroth law guarantees that if C is in equilibrium with A and C was in equilibrium with B at a known reading, then A is at that same known temperature — without ever needing to place A and B in direct contact with each other. That is the logical trick that makes a thermometer a valid measuring instrument at all.