Energy Transfers with the Surroundings
On a summer day in Mérida, the beach sand burns your feet at noon, but the ocean water is barely warm. A pot of boiling water stays on the stove and the thermometer never goes above $100\ ^\circ\text{C}$, even though the flame is still lit. An ice pack cools your injured knee without warming up. In all of these cases, energy passes from one place to another. In this section you will learn to measure how much energy is transferred and where it goes.
How Heat Travels
Energy always flows from hot to cold, until both bodies reach the same temperature. The energy that passes because of a temperature difference is called heat ($Q$) and is measured in joules (J). There are three ways to transfer it.
| Way | How it happens | Example |
|---|---|---|
| Conduction | Particles that vibrate more collide with neighboring particles and pass energy to them. It needs matter in contact. | The metal spoon gets hot in the coffee. |
| Convection | A hot fluid (less dense) rises and the cold fluid sinks; this is how energy moves along with the fluid. | The water in a boiling pot; the sea breeze. |
| Radiation | Electromagnetic waves. It does not need matter: it travels through a vacuum. | The Sun heats the Earth; a solar water heater. |
Specific Heat: $Q = m c \Delta T$
If you give the same amount of energy to $1\ \text{kg}$ of water and to $1\ \text{kg}$ of metal, the metal heats up much more. The reason is that each substance needs a different amount of energy to raise its temperature.
This is because temperature measures the average kinetic energy of the particles. Raising that energy in all the particles costs more or less depending on the substance and on how many particles there are (the mass).
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.
Latent Heat and Changes of State
If you heat ice with water, the thermometer reads $0\ ^\circ\text{C}$ for several minutes even though the stove is still on. Where does the energy go?
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 with Several Stages
Real problems combine stages: heating, changing state, heating again. The total heat is the sum of the heat in each stage. In mixtures with ice, before calculating a final temperature you must check whether the available heat is enough to melt all the ice.
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.
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.
First Law: $\Delta U = Q + W$
When you inflate a bicycle tire with a hand pump, the cylinder gets hot even though there is no flame: compressing the air increases its energy. How are heat and work combined?
Special cases: if the volume is constant (isochoric), $W = 0$ and $\Delta U = Q$. In an isothermal ideal gas, $\Delta U = 0$ and $Q = -W$: the heat absorbed is converted into work.
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.
A gas inside a cylinder absorbs heat $Q = +600\ \text{J}$ from a flame while the piston compresses it, doing work $W = +150\ \text{J}$ on it. The diagram shows the…
A gas inside a cylinder absorbs heat $Q = +600\ \text{J}$ from a flame while the piston compresses it, doing work $W = +150\ \text{J}$ on it. The diagram shows the energy transfers to the gas.
- a) Calculate the change in internal energy $\Delta U$ of the gas.
- b) Interpret the result: has the internal energy of the gas increased or decreased?
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 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.
Heat Engines and Efficiency
A car engine uses only part of the energy in the gasoline; the rest leaves through the exhaust and the radiator. Why can't it use all of 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.
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.
Go back to the beginning: the sand gets hotter than the sea because its specific heat is lower; the pot doesn't rise above $100\ ^\circ\text{C}$ because the energy goes into the change of state; and ice cools because it absorbs heat as it melts.