Ever wondered why a hot cup of tea eventually cools down, or how a car engine turns fuel into motion? That’s the first law of thermodynamics doing its quiet work.
💡 In Simple Words: Energy can’t be created or destroyed, it just moves around. When you add heat to a system, some of that energy raises its temperature (internal energy) and the rest can push things (do work).
What is the First Law of Thermodynamics?
The first law is basically the law of energy conservation for any system. In plain language it says:
Change in internal energy (ΔU) = Heat added to the system (Q) – Work done by the system (W).
Here internal energy is the total microscopic energy stored in the particles of a substance – think of it as the “wiggle‑energy” of atoms and molecules. Heat (Q) is energy transferred because of a temperature difference, like the warmth flowing from a hot pan to a cold spoon. Work (W) is energy transferred when a force moves something, such as a gas expanding and pushing a piston.
Why does it matter for Class 11 Physics?
CBSE and ISC exams love to test you on how you apply this rule to real situations – from a gas in a cylinder to a cup of coffee. Knowing the first law lets you connect heat, temperature, and mechanical work, which shows up in many problems.
Key Terms Explained
- System: The part of the universe you’re focusing on, like the gas inside a piston.
- Surroundings: Everything outside the system that can exchange energy with it.
- Isolated system: No heat or work crosses the boundary – its total energy stays constant.
- Closed system: Heat can cross the boundary but matter cannot.
How to Use the First Law – A Step‑by‑Step Guide
Whenever a question asks for a change in internal energy, heat, or work, follow these three steps:
- Identify what’s given: Q (heat added) and/or W (work done).
- Pay attention to signs: Heat added to the system is positive (+Q), heat removed is negative (‑Q). Work done *by* the system is positive (+W), work done *on* the system is negative (‑W).
- Plug into ΔU = Q – W and solve.
Worked Example 1 – Heating a Gas
A 2 mol ideal gas is heated at constant volume, absorbing 500 J of heat. No work is done because the volume doesn’t change. What’s the change in internal energy?
Given: Q = +500 J, W = 0 (constant volume → no expansion work).
Using the first law:
ΔU = Q – W = 500 J – 0 = 500 J. The gas’s internal energy rises by 500 J.
Worked Example 2 – Expanding Gas Does Work
A cylinder with a movable piston contains 1 mol of ideal gas at 300 K. The gas expands, doing 200 J of work on the piston, while it loses 150 J of heat to the surroundings. Find ΔU.
Here heat leaves the system, so Q = –150 J. Work is done *by* the gas, so W = +200 J.
ΔU = Q – W = (‑150 J) – (200 J) = ‑350 J. The internal energy drops because more energy left as work than came in as heat.
Common Mistakes to Avoid
- Mixing up sign conventions – remember “heat added” and “work done by the system” are positive.
- Forgetting that work can be negative when the surroundings compress the system.
- Assuming ΔU is always positive; it can be negative if the system loses more energy than it gains.
Quick Reference Table
| Scenario | Heat (Q) | Work (W) | ΔU = Q – W |
|---|---|---|---|
| Heating at constant volume | + (added) | 0 | Positive ΔU |
| Cooling while expanding | ‑ (removed) | + (done by system) | Usually negative ΔU |
| Compression with no heat exchange | 0 | ‑ (done on system) | Positive ΔU |
Energy Flow in a Simple System – Mermaid Flowchart
First Law in Everyday Life
Think of a rechargeable battery. When you plug it in, electrical energy (work) flows into the chemical system, raising its internal energy. When you use the phone, the stored internal energy turns back into electrical work and heat. The total amount of energy never disappears – it just changes form.
Where to Find More Practice
Many teachers upload a “first law of thermodynamics in physics pdf” that contains extra problems. Search for “physics first law of thermodynamics class 11 pdf” to get free worksheets.
📝 Likely Exam Questions
- Question: A gas absorbs 300 J of heat and does 120 J of work. Calculate the change in internal energy.
- Answer: ΔU = Q – W = 300 J – 120 J = 180 J.
- Question: In an adiabatic (no heat exchange) expansion, the gas does 250 J of work. What is ΔU?
- Answer: Q = 0, so ΔU = –W = –250 J (internal energy decreases).
- Question: A piston compresses a gas, work of 80 J is done on the system while 40 J of heat leaves the system. Find ΔU.
- Answer: Q = –40 J, W = –80 J (work on system is negative). ΔU = (‑40) – (‑80) = 40 J.
- Question: State the first law of thermodynamics in your own words.
- Answer: The total energy of a system changes only by the amount of heat added to it minus the work it does on its surroundings.