{"id":838,"date":"2026-08-20T19:37:51","date_gmt":"2026-08-20T19:37:51","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=838"},"modified":"2026-08-24T13:03:37","modified_gmt":"2026-08-24T13:03:37","slug":"first-law-of-thermodynamics","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/first-law-of-thermodynamics\/","title":{"rendered":"The First Law of Thermodynamics"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nThe first law of thermodynamics states that the change in a system&#8217;s internal energy equals the heat added to the system minus the work done by the system: \u0394U = Q \u2212 W. It is the law of conservation of energy applied to thermodynamic processes, meaning energy can change form \u2014 from heat to work or internal energy \u2014 but can never be created or destroyed.\n\n<\/p><\/div>\n\n<p>Every time you rub your hands together on a cold morning, you convert mechanical work into thermal energy \u2014 and your palms warm up. That warmth did not appear from nothing. It came from the effort your muscles supplied, following a rule that the universe enforces without exception.<\/p> <p>That rule is the first law of thermodynamics. It keeps track of every joule entering or leaving a system, and it guarantees that the books always balance. Understanding it unlocks the physics behind engines, refrigerators, weather, and your own metabolism.<\/p> <h2>What Is the First Law of Thermodynamics?<\/h2> <p>The first law of thermodynamics is the law of conservation of energy applied to systems that exchange heat and work with their surroundings. It says that when you add heat to a system or do work on it, the total energy the system gains must equal the total energy it receives \u2014 no more, no less.<\/p> <p>Think of a sealed container of gas sitting on a hotplate. You supply heat (Q) to the gas. Some of that energy makes the gas molecules jiggle faster, raising the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/heat-vs-temperature\/\">temperature \u2014 what physicists call internal energy<\/a>. Some of it may push a piston outward, doing work (W) on the surroundings. The first law simply insists that these two portions add up to the heat you put in.<\/p> <p>In other words, energy has only three places to go in a thermodynamic process: into internal energy, out as work, or in as heat. The first law is the accountant that tracks every transfer.<\/p> <p>This idea was not always obvious. Before the 1840s, scientists treated heat and mechanical work as separate, unrelated quantities. It took the careful experiments of James Prescott Joule \u2014 who measured the temperature rise of water churned by falling weights \u2014 to prove that work and heat are interchangeable forms of the same thing: <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-energy-in-physics\/\">energy<\/a>. NASA&#8217;s Glenn Research Center offers <a href=\"https:\/\/www1.grc.nasa.gov\/beginners-guide-to-aeronautics\/first-law-internal-energy\/\" target=\"_blank\" rel=\"noopener\">a clear walkthrough<\/a> of how Joule&#8217;s insight led to the modern definition of internal energy.<\/p> <h2>The First Law of Thermodynamics Formula<\/h2> <p>The first law of thermodynamics is written as:<\/p>\n\n<div class=\"pf-formula\">\u0394U = Q \u2212 W<\/div>\n\n<p>where:<\/p> <ul> <li><strong>\u0394U<\/strong> = change in internal energy of the system (joules, J)<\/li> <li><strong>Q<\/strong> = heat added to the system (joules, J)<\/li> <li><strong>W<\/strong> = work done <em>by<\/em> the system on its surroundings (joules, J)<\/li> <\/ul> <h3>Sign conventions<\/h3> <p>Getting the signs right trips up more students than the formula itself. Here is the convention used in most physics courses (the &#8220;physics&#8221; sign convention):<\/p> <div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\"> <table style=\"width:100%;border-collapse:collapse;\"> <thead> <tr style=\"background:#0A1628;color:#FAF6EE;\"> <th style=\"padding:10px 14px;text-align:left;\">Quantity<\/th> <th style=\"padding:10px 14px;text-align:left;\">Positive (+)<\/th> <th style=\"padding:10px 14px;text-align:left;\">Negative (\u2212)<\/th> <\/tr> <\/thead> <tbody> <tr style=\"background:#F5F2EA;color:#0A1628;\"> <td style=\"padding:10px 14px;\"><strong>Q (heat)<\/strong><\/td> <td style=\"padding:10px 14px;\">Heat flows <em>into<\/em> the system<\/td> <td style=\"padding:10px 14px;\">Heat flows <em>out of<\/em> the system<\/td> <\/tr> <tr style=\"background:#FAF6EE;color:#0A1628;\"> <td style=\"padding:10px 14px;\"><strong>W (work)<\/strong><\/td> <td style=\"padding:10px 14px;\">System does work <em>on<\/em> surroundings (expansion)<\/td> <td style=\"padding:10px 14px;\">Surroundings do work <em>on<\/em> system (compression)<\/td> <\/tr> <tr style=\"background:#F5F2EA;color:#0A1628;\"> <td style=\"padding:10px 14px;\"><strong>\u0394U<\/strong><\/td> <td style=\"padding:10px 14px;\">Internal energy increases (system heats up)<\/td> <td style=\"padding:10px 14px;\">Internal energy decreases (system cools down)<\/td> <\/tr> <\/tbody> <\/table> <\/div> <p>A common alternative in chemistry and some engineering texts writes the law as \u0394U = Q + W, where W is defined as work done <em>on<\/em> the system. Both forms say the same thing \u2014 just watch which sign convention the question uses.<\/p> <h3>What is internal energy?<\/h3> <p>Internal energy (U) is the total microscopic energy stored inside a system. For a gas, it includes the kinetic energy of every randomly moving molecule plus the potential energy from intermolecular forces. You cannot measure U directly, but you can measure <em>changes<\/em> in U \u2014 which is exactly what the first law does.<\/p> <p>For an ideal gas, internal energy depends only on temperature. Double the absolute temperature and you double U (for a monatomic ideal gas, U = <sup>3<\/sup>\u2044<sub>2<\/sub> nRT). That link between temperature and internal energy is why heating a sealed gas raises its temperature \u2014 all the added energy stays inside.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">First Law of Thermodynamics Lab<\/span><\/div><div class=\"pf-sim-slot-body\">\n\n<style> .pf-sim-frame{ width:100%; border:none; height:600px } @media(max-width:760px){ .pf-sim-frame{ height:1000px } } <\/style> <iframe src=\"\/labs\/first-law-of-thermodynamics.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"> <\/iframe> <\/div><\/div> <h2>How the First Law Works in 4 Thermodynamic Processes<\/h2> <p>The first law applies to every thermodynamic process, but its three terms \u2014 Q, W, and \u0394U \u2014 partition differently depending on what is held constant. Mastering these four standard processes is the fastest way to see the law in action.<\/p> <h3>1. Isothermal process (constant temperature)<\/h3> <p>Temperature stays fixed, so for an ideal gas \u0394U = 0. The first law reduces to Q = W: every joule of heat that enters is immediately spent as work expanding the gas. Slow expansion of a gas in contact with a heat reservoir is the textbook example.<\/p> <h3>2. Isobaric process (constant pressure)<\/h3> <p>Pressure does not change, so the work done by the gas is simply W = P\u0394V. The first law becomes \u0394U = Q \u2212 P\u0394V. Heating water in an open saucepan \u2014 where atmospheric pressure stays constant \u2014 is an everyday isobaric process.<\/p> <h3>3. Isochoric process (constant volume)<\/h3> <p>Volume is locked, so the gas does no expansion work: W = 0. The first law simplifies to \u0394U = Q \u2014 all the heat you add goes straight into internal energy, raising the temperature. A rigid, sealed pressure cooker before the valve opens behaves this way.<\/p> <h3>4. Adiabatic process (no heat transfer)<\/h3> <p>The system is perfectly insulated, so Q = 0. The first law gives \u0394U = \u2212W. If the gas expands and does positive work, its internal energy drops and it cools; if compressed, it heats up. The rapid compression stroke in a diesel engine \u2014 hot enough to ignite fuel without a spark plug \u2014 is a dramatic adiabatic process.<\/p> <div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\"> <table style=\"width:100%;border-collapse:collapse;\"> <thead> <tr style=\"background:#0A1628;color:#FAF6EE;\"> <th style=\"padding:10px 14px;text-align:left;\">Process<\/th> <th style=\"padding:10px 14px;text-align:left;\">Constraint<\/th> <th style=\"padding:10px 14px;text-align:left;\">Simplified first law<\/th> <th style=\"padding:10px 14px;text-align:left;\">Everyday example<\/th> <\/tr> <\/thead> <tbody> <tr style=\"background:#F5F2EA;color:#0A1628;\"> <td style=\"padding:10px 14px;\">Isothermal<\/td> <td style=\"padding:10px 14px;\">T = constant, \u0394U = 0<\/td> <td style=\"padding:10px 14px;\">Q = W<\/td> <td style=\"padding:10px 14px;\">Slow tyre inflation in contact with room air<\/td> <\/tr> <tr style=\"background:#FAF6EE;color:#0A1628;\"> <td style=\"padding:10px 14px;\">Isobaric<\/td> <td style=\"padding:10px 14px;\">P = constant<\/td> <td style=\"padding:10px 14px;\">\u0394U = Q \u2212 P\u0394V<\/td> <td style=\"padding:10px 14px;\">Boiling water in an open pan<\/td> <\/tr> <tr style=\"background:#F5F2EA;color:#0A1628;\"> <td style=\"padding:10px 14px;\">Isochoric<\/td> <td style=\"padding:10px 14px;\">V = constant, W = 0<\/td> <td style=\"padding:10px 14px;\">\u0394U = Q<\/td> <td style=\"padding:10px 14px;\">Heating gas in a rigid sealed container<\/td> <\/tr> <tr style=\"background:#FAF6EE;color:#0A1628;\"> <td style=\"padding:10px 14px;\">Adiabatic<\/td> <td style=\"padding:10px 14px;\">Q = 0<\/td> <td style=\"padding:10px 14px;\">\u0394U = \u2212W<\/td> <td style=\"padding:10px 14px;\">Rapid compression in a diesel engine<\/td> <\/tr> <\/tbody> <\/table> <\/div> <h2>Real-World Examples of the First Law of Thermodynamics<\/h2> <p>The first law of thermodynamics operates everywhere energy changes hands \u2014 not just inside laboratory cylinders. Here are five situations you can connect to \u0394U = Q \u2212 W right now.<\/p> <h3>1. A car engine<\/h3> <p>Burning fuel releases heat (Q &gt; 0). Part of that energy does work pushing pistons (W &gt; 0), and the remainder heats the engine block and exhaust gases (\u0394U and waste heat to the surroundings). The first law guarantees that the work output plus the thermal energy carried away equals the chemical energy released.<\/p> <h3>2. A refrigerator<\/h3> <p>The compressor does work on the refrigerant gas (W &lt; 0 from the gas&#8217;s perspective, since work is done <em>on<\/em> it). That compressed gas then dumps heat out the back of the fridge (Q &lt; 0, heat leaving). The first law explains why the coils at the back feel warm \u2014 the energy your food lost plus the electrical work the motor did all has to go somewhere.<\/p> <h3>3. A bicycle pump<\/h3> <p>Push the handle down quickly and the air inside the pump warms up. You did work on the gas (W &lt; 0), there was not enough time for heat to escape (roughly adiabatic, Q \u2248 0), so the internal energy rose: \u0394U = \u2212W &gt; 0. The nozzle gets hot to the touch.<\/p> <h3>4. Melting an ice cube<\/h3> <p>Heat flows in from the surroundings (Q &gt; 0) and breaks molecular bonds rather than raising temperature. The ice does negligible work (volume barely changes), so nearly all of Q becomes an increase in internal (potential) energy. This is also how <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/latent-heat\/\">latent heat<\/a> fits into the first law.<\/p> <h3>5. Your body<\/h3> <p>Food provides chemical energy (analogous to Q). Your muscles convert some of it into mechanical work (W), and the rest becomes body heat (\u0394U and thermal radiation). On a cold day your body &#8220;wastes&#8221; more energy as heat to maintain 37 \u00b0C \u2014 perfectly consistent with the first law&#8217;s energy balance.<\/p> <figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/first-law-of-thermodynamics-heat-q-enters-system-work.webp\" width=\"1400\" height=\"720\" alt=\"Diagram showing the first law of thermodynamics: heat Q enters a system, work W leaves, and internal energy \u0394U changes inside\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:700px;display:block;margin:0 auto;\" \/><\/figure> <p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;\">First law of thermodynamics energy-flow diagram: heat (Q) enters the system, work (W) leaves, and the balance is stored as internal energy (\u0394U).<\/p> <h2>Common Misconceptions About the First Law of Thermodynamics<\/h2> <p>Even students who can write \u0394U = Q \u2212 W from memory sometimes carry hidden misunderstandings. Here are four to catch early.<\/p> <h3>1. &#8220;Heat and temperature are the same thing&#8221;<\/h3> <p>They are not. Heat (Q) is energy <em>in transit<\/em> between objects at different temperatures. Temperature is a measure of the average kinetic energy of particles. You can add enormous amounts of heat to a substance \u2014 ice melting at 0 \u00b0C, for instance \u2014 without its temperature rising at all. Confusing the two leads to garbled first-law calculations. Our article on <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/heat-vs-temperature\/\">heat versus temperature<\/a> unpacks this in detail.<\/p> <h3>2. &#8220;Work &#8216;uses up&#8217; energy&#8221;<\/h3> <p>Work does not destroy energy \u2014 it transfers it from one system to another. When a gas expands and pushes a piston, the gas loses internal energy but the piston (and whatever it is connected to) gains kinetic or potential energy. The total remains constant.<\/p> <h3>3. &#8220;A system in thermal equilibrium has zero internal energy&#8221;<\/h3> <p>Internal energy is never zero (molecules are always moving above absolute zero). Equilibrium simply means \u0394U = 0 because no net heat or work is flowing. The gas still carries a huge store of kinetic energy inside \u2014 it just is not changing.<\/p> <h3>4. &#8220;The first law forbids heat from flowing from cold to hot&#8221;<\/h3> <p>That restriction belongs to the <em>second<\/em> law of thermodynamics. The first law only says energy is conserved; it has nothing to say about direction. A refrigerator moves heat from cold to hot \u2014 perfectly allowed by the first law \u2014 so long as external work is supplied. The <a href=\"https:\/\/en.wikipedia.org\/wiki\/First_law_of_thermodynamics\" target=\"_blank\" rel=\"noopener\">broader mathematical framework<\/a> behind the law makes this distinction precise: energy conservation constrains the magnitude of transfers, not their spontaneity.<\/p> <h2>How the First Law Connects to Other Physics Concepts<\/h2> <p>The first law of thermodynamics does not stand alone \u2014 it is the energy-conservation thread that runs through almost every branch of thermal physics.<\/p> <p>Start with the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/ideal-gas-law\/\">ideal gas law<\/a> (PV = nRT). For an ideal gas, internal energy depends only on temperature, so combining PV = nRT with \u0394U = Q \u2212 W lets you predict exactly how pressure, volume, and temperature shift in each of the four processes above.<\/p> <p>Move to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/specific-heat-capacity\/\">specific heat capacity<\/a> and the relationship becomes even more practical: Q = mc\u0394T tells you how much heat is needed, while the first law tells you where that heat ends up \u2014 as a temperature change, as expansion work, or as both.<\/p> <p>The first law also sits inside the broader family of the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/laws-of-thermodynamics\/\">four laws of thermodynamics<\/a>. While the zeroth law defines temperature and the first law conserves energy, the second law \u2014 via concepts like <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/carnot-efficiency\/\">Carnot efficiency<\/a> \u2014 adds a direction: it limits <em>how much<\/em> of Q you can convert to useful work. Together, the first and second laws explain why no engine can ever be 100 % efficient.<\/p> <p>Finally, the first law reaches into <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/conduction-convection-radiation\/\">heat transfer<\/a>. Conduction, convection, and radiation are simply the <em>mechanisms<\/em> by which Q crosses a system boundary \u2014 the first law then decides how that incoming energy is split between \u0394U and W.<\/p> <h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A gas in a sealed rigid container absorbs 500 J of heat. How much does its internal energy change?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Identify constraints. The container is rigid, so volume cannot change and W = 0.<\/p>\n<p>Step 2: Apply the first law: \u0394U = Q \u2212 W = 500 J \u2212 0 = 500 J.<\/p>\n<p><strong>Answer: \u0394U = 500 J (the internal energy increases by 500 J).<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A gas expands at constant pressure and does 200 J of work on its surroundings. If 600 J of heat is added to the gas, what is \u0394U?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: List knowns. Q = +600 J (heat in), W = +200 J (work done by the gas).<\/p>\n<p>Step 2: Apply \u0394U = Q \u2212 W = 600 J \u2212 200 J = 400 J.<\/p>\n<p><strong>Answer: \u0394U = 400 J.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">During an adiabatic compression, 350 J of work is done on a gas. Find \u0394U.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Adiabatic means Q = 0.<\/p>\n<p>Step 2: Work is done <em>on<\/em> the gas, so by the physics convention W = \u2212350 J (negative because the surroundings do the work).<\/p>\n<p>Step 3: \u0394U = Q \u2212 W = 0 \u2212 (\u2212350) = +350 J.<\/p>\n<p><strong>Answer: \u0394U = +350 J (the gas heats up).<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">An ideal gas undergoes an isothermal expansion at 300 K. It absorbs 1,200 J of heat from a reservoir. How much work does the gas do?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Isothermal for an ideal gas means \u0394U = 0 (internal energy depends only on T).<\/p>\n<p>Step 2: \u0394U = Q \u2212 W gives 0 = 1,200 J \u2212 W.<\/p>\n<p>Step 3: W = 1,200 J.<\/p>\n<p><strong>Answer: W = 1,200 J \u2014 the gas does 1,200 J of work on the piston.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A system releases 800 J of heat to its surroundings while 300 J of work is done on the system. What is \u0394U?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Heat leaves the system, so Q = \u2212800 J.<\/p>\n<p>Step 2: Work is done on the system, so W = \u2212300 J (the system does \u2212300 J of work on the surroundings).<\/p>\n<p>Step 3: \u0394U = Q \u2212 W = (\u2212800) \u2212 (\u2212300) = \u2212800 + 300 = \u2212500 J.<\/p>\n<p><strong>Answer: \u0394U = \u2212500 J (the internal energy decreases by 500 J).<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">In one full cycle of a heat engine, the working gas absorbs 4,000 J of heat from the hot reservoir and rejects 2,500 J to the cold reservoir. How much net work does the engine produce per cycle?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Over a complete cycle the gas returns to its initial state, so \u0394U = 0.<\/p>\n<p>Step 2: Net heat absorbed: Q<sub>net<\/sub> = Q<sub>in<\/sub> \u2212 Q<sub>out<\/sub> = 4,000 J \u2212 2,500 J = 1,500 J.<\/p>\n<p>Step 3: \u0394U = Q<sub>net<\/sub> \u2212 W gives 0 = 1,500 J \u2212 W, so W = 1,500 J.<\/p>\n<p><strong>Answer: W = 1,500 J of net work per cycle.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A monatomic ideal gas (n = 2.0 mol) is heated at constant volume from 300 K to 500 K. Calculate Q, W, and \u0394U. (Use C_v = 3R\/2 for a monatomic ideal gas; R = 8.314 J\/(mol\u00b7K).)<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Constant volume means W = 0.<\/p>\n<p>Step 2: \u0394U = nC<sub>v<\/sub>\u0394T = 2.0 mol \u00d7 (3 \u00d7 8.314 \/ 2) J\/(mol\u00b7K) \u00d7 (500 \u2212 300) K.<\/p>\n<p>Step 3: C<sub>v<\/sub> = 12.471 J\/(mol\u00b7K). \u0394U = 2.0 \u00d7 12.471 \u00d7 200 = 4,988 J \u2248 4,990 J.<\/p>\n<p>Step 4: Since W = 0, Q = \u0394U = 4,990 J.<\/p>\n<p><strong>Answer: Q \u2248 4,990 J, W = 0 J, \u0394U \u2248 4,990 J.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">A gas expands from 0.0020 m^3 to 0.0060 m^3 at a constant pressure of 150 kPa. During the expansion, 900 J of heat is added. Find W and \u0394U.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Work at constant pressure: W = P\u0394V = 1.5 \u00d7 10<sup>5<\/sup> Pa \u00d7 (6.0 \u00d7 10<sup>-3<\/sup> \u2212 2.0 \u00d7 10<sup>-3<\/sup>) m<sup>3<\/sup>.<\/p>\n<p>Step 2: W = 1.5 \u00d7 10<sup>5<\/sup> \u00d7 4.0 \u00d7 10<sup>-3<\/sup> = 600 J.<\/p>\n<p>Step 3: \u0394U = Q \u2212 W = 900 J \u2212 600 J = 300 J.<\/p>\n<p><strong>Answer: W = 600 J, \u0394U = 300 J.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is the first law of thermodynamics in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe first law says energy can change form but cannot be created or destroyed. For any thermodynamic system, the change in internal energy equals the heat added minus the work the system does: \u0394U = Q \u2212 W. It is conservation of energy applied to heat and work.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between the first and second law of thermodynamics?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe first law conserves energy \u2014 it tells you the total is constant. The second law restricts direction \u2014 it says heat flows spontaneously only from hot to cold, and no engine can convert all heat into work. The first law is a balance sheet; the second law is a one-way sign.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why is internal energy a state function but heat and work are not?<\/summary><div class=\"pf-faq-item-answer\">\n\nInternal energy depends only on the current state (temperature, pressure, volume) of a system. Heat and work depend on the path taken between two states \u2014 you can reach the same final state with different combinations of Q and W. That makes Q and W process quantities, not state functions.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does the first law of thermodynamics apply to living organisms?<\/summary><div class=\"pf-faq-item-answer\">\n\nYes. Your body obeys the first law: chemical energy from food equals the work your muscles do plus the heat your body radiates plus any change in stored energy (fat, glycogen). No biological process creates energy from nothing.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What happens to the first law in a cyclic process?<\/summary><div class=\"pf-faq-item-answer\">\n\nIn a complete cycle the system returns to its starting state, so \u0394U = 0. The first law then gives Q<sub>net<\/sub> = W<sub>net<\/sub> \u2014 the net heat absorbed equals the net work output. This is the operating principle behind every heat engine.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Can the first law of thermodynamics be violated?<\/summary><div class=\"pf-faq-item-answer\">\n\nNo verified experiment has ever violated the first law. A perpetual-motion machine of the first kind \u2014 one that produces work with no energy input \u2014 would break it, and none has ever been built. The law is considered one of the most robust principles in all of physics.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Who discovered the first law of thermodynamics?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe first law emerged from the work of several scientists in the 1840s. James Prescott Joule demonstrated the mechanical equivalent of heat experimentally, while Julius Robert von Mayer and Hermann von Helmholtz independently formulated the general principle of energy conservation. Rudolf Clausius later formalised the law in its modern thermodynamic form.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>The first law of thermodynamics states that the change in a system&#8217;s internal energy equals the heat added minus the work done: \u0394U = Q \u2212 W. Explore 4 thermodynamic processes, 8 worked problems, and an interactive lab.<\/p>\n","protected":false},"author":1,"featured_media":839,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"class_list":["post-838","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-thermodynamics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/838","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=838"}],"version-history":[{"count":9,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/838\/revisions"}],"predecessor-version":[{"id":1422,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/838\/revisions\/1422"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/839"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=838"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=838"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=838"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}