Energy Transfers with the Surroundings
When the power goes out in Santo Domingo and the inverter runs out of charge, the refrigerator stops cooling and everything in the freezer starts to melt. How much energy does it take to melt an ice cube? And why does the refrigerator, when it is working, heat up the kitchen? In this section you will do energy balances with clear signs: what enters and what leaves a system.
How Heat Is Transferred
Heat is energy that passes from one body to another because they are at different temperatures, always from the hotter one to the colder one, until they reach thermal equilibrium (the zeroth law). There are three mechanisms:
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Conduction: energy passes by contact, from molecule to molecule (the metal spoon in the sancocho gets hot).
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Convection: a mass of hot fluid moves and carries the energy with it (the sea breeze, the water boiling in the pot).
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Radiation: energy travels as an electromagnetic wave, without needing matter (the Sun heating the sand in Punta Cana).
Thermal Expansion
When heated, almost all materials get a little bigger. The molecules vibrate with greater amplitude and, on average, move farther apart. That is why bridges and rails have expansion joints, and why the liquid in a thermometer rises when it is heated.
The relationship with heat is direct: when a body receives heat, its temperature rises ($Q = m\,c\,\Delta T$) and, with it, the size of the body.
The First Law with a Sign Convention
When you inflate a bicycle tire with a hand pump, the lower part of the pump gets hot even though it is not near any flame. You have done work on the air, and that increased its energy. How do heat and work combine to tell whether a gas gains or loses internal energy?
Some books write $\Delta U = Q - W$, with $W$ being the work done by the system. Here we always use $\Delta U = Q + W$.
Particular processes
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Isochoric (constant volume): $\Delta V = 0$, so $W = -p\,\Delta V = 0$ and $\Delta U = Q$.
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Isothermal (constant temperature) in an ideal gas: $\Delta U = 0$, so $Q = -W$.
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Isobaric (constant pressure): $W = -p\,\Delta V$.
In thermodynamics we will use the following sign convention: $Q > 0$ when the system absorbs heat from the surroundings and $Q < 0$ when it gives off heat; $W > 0$ when…
In thermodynamics we will use the following sign convention: $Q > 0$ when the system absorbs heat from the surroundings and $Q < 0$ when it gives off heat; $W > 0$ when the surroundings do work on the system (compression) and $W < 0$ when the system does work on the surroundings (expansion). With this convention, the first law is written as $\Delta U = Q + W$.
Write with its sign the value of $Q$ or $W$ in each case:
- a) The system absorbs $400\ \text{J}$ of heat.
- b) The system gives off $250\ \text{J}$ of heat to the surroundings.
- c) The surroundings compress the system, doing $100\ \text{J}$ of work.
- d) The system expands and does $150\ \text{J}$ of work on the piston.
The hot water in a home heating circuit passes through a radiator and gives off $500\ \text{J}$ of heat to the room, without any work being done. a) Write the values of…
The hot water in a home heating circuit passes through a radiator and gives off $500\ \text{J}$ of heat to the room, without any work being done.
- a) Write the values of $Q$ and $W$ with their sign according to the standard convention.
- b) Calculate $\Delta U$ of the water in that section and interpret its sign.
State the first law of thermodynamics in your own words. a) Explain what each of the three terms in the expression $\Delta U = Q + W$ represents. b) Reason why this law…
State the first law of thermodynamics in your own words.
- a) Explain what each of the three terms in the expression $\Delta U = Q + W$ represents.
- b) Reason why this law is, at heart, the principle of conservation of energy applied to thermodynamic systems.
A gas gives off $300\ \text{J}$ of heat to the surroundings while a piston compresses it, doing work of $150\ \text{J}$ on it. a) Write $Q$ and $W$ with their sign…
A gas gives off $300\ \text{J}$ of heat to the surroundings while a piston compresses it, doing work of $150\ \text{J}$ on it.
- a) Write $Q$ and $W$ with their sign according to the standard convention.
- b) Calculate $\Delta U$ and interpret the result: despite the compression, does the gas gain or lose internal energy?
- c) For a gas whose internal energy increases with temperature, predict whether its temperature will have risen or fallen.
A closed pressure cooker, before its valve opens, maintains the volume of steam practically constant while receiving heat $Q = +4000\ \text{J}$ from the hob. a) Justify…
A closed pressure cooker, before its valve opens, maintains the volume of steam practically constant while receiving heat $Q = +4000\ \text{J}$ from the hob.
- a) Justify why the work exchanged is zero in this stage (think about the expression $W = -p \times \Delta V$).
- b) Calculate the change in internal energy $\Delta U$ of the contents in that stage.
- c) Generalise the result by completing the sentence: "in any isochoric process, the change in internal energy is equal to $\dots$".
An ideal gas expands very slowly inside a thermostatic bath (isothermal process) and absorbs heat $Q = +500\ \text{J}$ from the bath. Remember that, in an ideal gas,…
An ideal gas expands very slowly inside a thermostatic bath (isothermal process) and absorbs heat $Q = +500\ \text{J}$ from the bath. Remember that, in an ideal gas, internal energy depends only on temperature.
- a) State the value of $\Delta U$ in this process and justify your answer.
- b) Calculate the work $W$ exchanged, with its sign, and interpret it.
- c) Explain, using the energy balance, what the absorbed heat has been converted into.
A gas expands at constant pressure $p = 1\times10^{5}\ \text{Pa}$ from $V_1 = 1\times10^{-3}\ \text{m}^3$ to $V_2 = 3\times10^{-3}\ \text{m}^3$ and, during the…
A gas expands at constant pressure $p = 1\times10^{5}\ \text{Pa}$ from $V_1 = 1\times10^{-3}\ \text{m}^3$ to $V_2 = 3\times10^{-3}\ \text{m}^3$ and, during the expansion, absorbs heat $Q = +700\ \text{J}$ from a burner.
- a) Calculate the work $W = -p \times \Delta V$ with its sign.
- b) Calculate the change in internal energy $\Delta U$ of the gas.
- c) Verify that the heat absorbed is divided between the increase in internal energy and the work done by the gas.
A gas absorbs heat $Q = +1000\ \text{J}$ from a burner and its internal energy increases by $600\ \text{J}$ while pushing the piston of a mechanism. a) Calculate the…
A gas absorbs heat $Q = +1000\ \text{J}$ from a burner and its internal energy increases by $600\ \text{J}$ while pushing the piston of a mechanism.
- a) Calculate the work $W$ exchanged by the gas, including its sign, and state how much work the gas has done.
- b) The piston experiences friction, so out of that work only $300\ \text{J}$ is delivered as useful work to the external mechanism. Explain what happened to the rest.
- c) Verify that the balance $\Delta U = Q + W$ still holds and explain why friction does not "break" the conservation of energy even though it reduces useful work.
Specific Heat: $Q = m \times c \times \Delta T$
An electric kettle heats $m = 2\ \text{kg}$ of water and raises its temperature by $\Delta T = 10\ \celsius$. The specific heat capacity of water is…
An electric kettle heats $m = 2\ \text{kg}$ of water and raises its temperature by $\Delta T = 10\ \celsius$. The specific heat capacity of water is $c = 4180\ \text{J}/(\text{kg}\times\celsius)$.
Calculate the heat absorbed by the water using the expression $Q = m \times c \times \Delta T$.
The specific heat capacity of water is $c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$ and that of ethanol is…
The specific heat capacity of water is $c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$ and that of ethanol is $c_{\text{etanol}} = 2400\ \text{J}/(\text{kg}\times\celsius)$.
- a) Explain in your own words what the value $4180\ \text{J}/(\text{kg}\times\celsius)$ means: how many joules are needed and to achieve what, exactly.
- b) For equal masses and the same temperature increase, reason using formula $Q = m \times c \times \Delta T$ which of the two substances requires more heat.
In the laboratory, $750\ \text{mL}$ of water is heated from $20\ \celsius$ to $60\ \celsius$. Data: the density of water is $1\ \text{kg/L}$ and its specific heat…
In the laboratory, $750\ \text{mL}$ of water is heated from $20\ \celsius$ to $60\ \celsius$. Data: the density of water is $1\ \text{kg/L}$ and its specific heat capacity is $c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$.
- a) Express the mass of water in kilograms and calculate the temperature change.
- b) Calculate the heat absorbed by the water and express it in standard form.
Heat $Q = 11\,500\ \text{J}$ is supplied to a metal block of $m = 0.5\ \text{kg}$, initially at $20\ \celsius$, and its temperature rises to $70\ \celsius$. a) Rearrange…
Heat $Q = 11\,500\ \text{J}$ is supplied to a metal block of $m = 0.5\ \text{kg}$, initially at $20\ \celsius$, and its temperature rises to $70\ \celsius$.
- a) Rearrange the formula $Q = m \times c \times \Delta T$ for the specific heat capacity $c$.
- b) Calculate its value in $\text{J}/(\text{kg}\times\celsius)$.
- c) By checking the values $c_{\text{plomo}} = 130\ \text{J}/(\text{kg}\times\celsius)$, $c_{\text{hierro}} = 460\ \text{J}/(\text{kg}\times\celsius)$ and $c_{\text{aluminio}} = 900\ \text{J}/(\text{kg}\times\celsius)$, identify which metal it could be.
On a clear day, the Sun transfers the same energy $Q = 4180\ \text{J}$ to $1\ \text{kg}$ of sea water ($c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$) and…
On a clear day, the Sun transfers the same energy $Q = 4180\ \text{J}$ to $1\ \text{kg}$ of sea water ($c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$) and to $1\ \text{kg}$ of beach sand ($c_{\text{arena}} = 836\ \text{J}/(\text{kg}\times\celsius)$).
- a) Calculate the temperature increase of the water.
- b) Calculate the temperature increase of the sand.
- c) Use the results to explain why the sand is burning hot at midday while the sea water is barely lukewarm.
- d) Relate this to daytime sea breezes: reason over which surface the air heats up first and in which direction the breeze blows.
Latent Heat and Changes of State
If you heat a mixture of water and ice, the thermometer reads 0 °C for several minutes until the last piece of ice disappears, even though the flame is still on. Why doesn't the temperature rise while the two phases coexist?
An ice pack containing $m = 0.5\ \text{kg}$ of ice at $0\ \celsius$ is applied to a sports injury until all the ice melts, remaining at $0\ \celsius$ throughout the…
An ice pack containing $m = 0.5\ \text{kg}$ of ice at $0\ \celsius$ is applied to a sports injury until all the ice melts, remaining at $0\ \celsius$ throughout the process. The latent heat of fusion of ice is $L_{\text{f}} = 3.34\times10^{5}\ \text{J/kg}$.
- a) Calculate the heat absorbed by the ice using $Q = m \times L_{\text{f}}$ and express the result in standard form.
- b) State where that heat comes from and, therefore, explain the cooling effect on the treated area.
A pot of boiling water remains on the hob, but the thermometer reads $100\ \celsius$ throughout the process, even though the flame continues to supply energy. The graph…
A pot of boiling water remains on the hob, but the thermometer reads $100\ \celsius$ throughout the process, even though the flame continues to supply energy. The graph shows how the temperature of the water changes over time.
- a) Does the water continue to receive energy while it is boiling? Justify your answer.
- b) If the temperature does not rise, what is that energy being used for at a microscopic level?
- c) Explain, using the concept of latent heat, why the temperature remains constant during a change of state.
In an industrial boiler, $m = 0.2\ \text{kg}$ of water is heated from $80\ \celsius$ to $100\ \celsius$ and then that mass is completely vaporised at $100\ \celsius$.…
In an industrial boiler, $m = 0.2\ \text{kg}$ of water is heated from $80\ \celsius$ to $100\ \celsius$ and then that mass is completely vaporised at $100\ \celsius$. Data: $c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$ and $L_{\text{v}} = 2.26\times10^{6}\ \text{J/kg}$.
- a) Calculate the heat required for the first stage (heating the water).
- b) Calculate the heat required for complete vaporisation using $Q = m \times L_{\text{v}}$.
- c) Calculate the total heat of the process and comment on which of the two stages turned out to be much more costly in terms of energy.
Energy Balances in Mixed Problems
We want to convert $m = 0.5\ \text{kg}$ of ice at $-10\ \celsius$ into liquid water at $40\ \celsius$. Data:…
We want to convert $m = 0.5\ \text{kg}$ of ice at $-10\ \celsius$ into liquid water at $40\ \celsius$. Data: $c_{\text{hielo}} = 2100\ \text{J}/(\text{kg}\times\celsius)$, $c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$ and $L_{\text{f}} = 3.34\times10^{5}\ \text{J/kg}$.
- a) Calculate the heat required to heat the ice from $-10\ \celsius$ to $0\ \celsius$.
- b) Calculate the heat required to melt all the ice at $0\ \celsius$.
- c) Calculate the heat required to heat the resulting water from $0\ \celsius$ to $40\ \celsius$.
- d) Calculate the total heat of the process, express it in standard form and draw the graph of temperature as a function of heat supplied in your notebook, marking the different stages.
In an insulated container there is $0.3\ \text{kg}$ of water at $60\ \celsius$ and $0.3\ \text{kg}$ of ice at $0\ \celsius$ is added. Data:…
In an insulated container there is $0.3\ \text{kg}$ of water at $60\ \celsius$ and $0.3\ \text{kg}$ of ice at $0\ \celsius$ is added. Data: $c_{\text{agua}} = 4180\ \text{J}/(\text{kg}\times\celsius)$ and $L_{\text{f}} = 3.34\times10^{5}\ \text{J/kg}$.
- a) Calculate the maximum heat that the hot water can give off if it cooled down to $0\ \celsius$.
- b) Calculate the heat needed to melt all the ice.
- c) Comparing both results, predict whether all the ice will melt and deduce the final temperature of the mixture.
- d) Calculate the mass of ice that melts and the mass that remains unmelted.
Heat Engines, Refrigerators, and Entropy
A car engine burns fuel to move the car, but more than half of that energy leaves through the exhaust and the radiator. Why can't 100% be used? The same thing happens in a thermoelectric plant like Punta Catalina: part of the energy is given off to the cooling water.
Heat engines have a long history. The steam engine burns coal or wood to boil water; the steam pushes a piston and produces work (that is how trains and the old sugar mills worked). The internal combustion engines of cars and motoconchos burn gasoline inside the cylinder and the hot gas pushes the piston. In all of them, the scheme is the same.
The second law and entropy
Heat is transferred at the expense of the chaotic, disordered agitation of many molecules (internal energy), while work is energy organized in one direction. Converting all the absorbed heat into work, without giving any to a cold reservoir, would make the total entropy of the universe decrease (the hot reservoir loses entropy with no compensation), which the second law forbids. Entropy is the measure of that disorder, and the second law says that in a spontaneous process the total entropy of the universe does not decrease. That is why heat flows only from hot to cold, and why no heat engine has an efficiency of 100%. Here it is just a concept: you do not have to calculate it.
In a thermal power plant, the engine driving the generator absorbs heat $Q_1 = 2000\ \text{J}$ per cycle from the hot reservoir (the boiler) and gives off heat…
In a thermal power plant, the engine driving the generator absorbs heat $Q_1 = 2000\ \text{J}$ per cycle from the hot reservoir (the boiler) and gives off heat $Q_2 = 1200\ \text{J}$ to the cold reservoir (the cooling water). In a complete cycle, the change in internal energy of the engine is zero. The heat engine diagram is shown in the figure.
- a) Calculate the useful work $W_{\text{useful}}$ delivered by the engine in each cycle from the energy balance.
- b) The efficiency of a heat engine is defined as $\eta = W_{\text{useful}}/Q_1$. Calculate the efficiency of this engine and express it as a percentage.
- c) Interpret the result by completing the sentence: "for every $100\ \text{J}$ of heat absorbed, the engine uses $\dots$ and transfers $\dots$ to the cold reservoir".
A heat engine undergoes closed cycles. In each cycle it absorbs heat $Q_1 = +3000\ \text{J}$ from the hot reservoir and delivers work of $900\ \text{J}$ to the…
A heat engine undergoes closed cycles. In each cycle it absorbs heat $Q_1 = +3000\ \text{J}$ from the hot reservoir and delivers work of $900\ \text{J}$ to the surroundings, that is, $W = -900\ \text{J}$ with the standard sign convention. The diagram of the engine is shown in the figure.
- a) Justify why the change in internal energy of the engine in a complete cycle is zero.
- b) Apply $\Delta U = Q + W$ to the cycle and calculate the net heat $Q$ exchanged per cycle.
- c) Taking into account that $Q = Q_1 + Q_2$, calculate the heat $Q_2$ exchanged with the cold reservoir, with its sign, and interpret it.
- d) Calculate the efficiency $\eta = W_{\text{useful}}/Q_1$ of the engine.
A refrigerator operates as a heat engine in reverse: in each cycle, it extracts heat $Q_2 = +600\ \text{J}$ from the inside and the electric motor does work…
A refrigerator operates as a heat engine in reverse: in each cycle, it extracts heat $Q_2 = +600\ \text{J}$ from the inside and the electric motor does work $W = +200\ \text{J}$ on the system. In a complete cycle, $\Delta U = 0$. The diagram of the refrigerator is shown in the figure.
- a) Apply the first law of thermodynamics to the cycle and calculate the net heat $Q$ exchanged.
- b) Calculate the heat $Q_1$ that the refrigerator releases to the kitchen, including its sign, and interpret it: why is it greater than the heat extracted from the inside?
- c) Use the result to explain why leaving the refrigerator door open does not serve to cool down the kitchen, but actually ends up heating it up.
Two heat engines work between the same reservoirs. Engine A absorbs $Q_1 = 5000\ \text{J}$ from the hot reservoir and delivers useful work of $1500\ \text{J}$. Engine B…
Two heat engines work between the same reservoirs. Engine A absorbs $Q_1 = 5000\ \text{J}$ from the hot reservoir and delivers useful work of $1500\ \text{J}$. Engine B absorbs the same $5000\ \text{J}$ and releases heat of $2500\ \text{J}$ to the cold reservoir.
- a) Calculate the efficiency of engine A.
- b) Calculate the useful work of engine B from its energy balance and then its efficiency.
- c) Compare both engines: which one makes better use of the fuel's heat?
- d) An inventor claims that they can build an engine with an efficiency of $100\ \%$, which releases no heat to the cold reservoir. Using the energy balance and the concept that every real heat engine must release heat to a cold reservoir, explain why such an engine is not possible.
A thermal power station receives $5\times10^{6}\ \text{J}$ of energy from combustion per unit of time considered and delivers $1.5\times10^{6}\ \text{J}$ in the form of…
A thermal power station receives $5\times10^{6}\ \text{J}$ of energy from combustion per unit of time considered and delivers $1.5\times10^{6}\ \text{J}$ in the form of electrical work; the rest is released as heat to the cooling water of a river.
- a) Calculate the heat released to the cooling water.
- b) Calculate the current efficiency of the power station.
- c) Following a technological upgrade, the power station achieves an efficiency of $45\ \%$ with the same combustion input. Calculate the new work delivered and the new heat released to the river.
- d) Using your results to support your answer, explain why increasing the efficiency of power stations reduces the environmental impact of thermal pollution on river ecosystems.
Back to the beginning: the refrigerator heats the kitchen because it gives off more heat than it extracts, and the ice in the freezer absorbs $Q = m\,L_f$ to melt without rising above 0 °C.