riaology — where curiosity flows

Foundations of thermodynamics

Small systems. Big ideas.

01 · Calculate

How much energy leaves the coffee?

A 0.250 kg coffee cools from 80°C to 40°C. Use $c = 4.18 \mathrm{kJ}$ /(kg·K). Neglect evaporation, work, and changes in kinetic and potential energy. Find the energy change of the coffee.

A small hint

The temperature change is final minus initial. Keep kJ and J separate.

Explore the visual solution

Coffee energy change-41.8 kJ

Energy transferred to surroundings41.8 kJ

What if you change it?Final temperature40 °C

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Choose the coffee as a fixed-mass system. With work and motion neglected, $Q = \Delta U$ .

Carry this forward. The temperature drop gives the energy change; it does not by itself tell you how long cooling takes.

02 · Reason

A room with a running heater is cooling.

A heater supplies 1,000 W to a room that is losing 1,200 W. The effective room heat capacity is 60 kJ/K. Find its instantaneous temperature-change rate.

A small hint

Find the net power first, then divide by heat capacity.

Explore the visual solution

Heater input1,000 W

Heat loss1,200 W

Instantaneous temperature-change rate-0.2 K/min

Energy storage decreases; the room cools.

What if you change it?Heat loss at this instant1,200 W

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Energy accumulates at $1,000 - 1,200 = -200 \mathrm{J}$ /s.

Carry this forward. This is an instantaneous rate; losses may change as the room temperature changes.

03 · Challenge

Can this pump deliver the advertised flow?

A pump must raise pressure by 100 kPa at 12 L/min. Its shaft input is 15 W. Ignore elevation and speed changes. Is the claim possible at steady state, even without losses?

A small hint

Hydraulic power is pressure $rise \times volume$ flow rate. Convert liters per minute to m³/s.

Explore the visual solution

Required hydraulic power20 W

Available shaft power15 W

Energy feasibilityInsufficient input

At this flow, ideal hydraulic output is 20 W. Other constraints and losses still matter.

What if you change it?Required flow12 L/min

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Convert 12 L/min to 0.000200 m³/s.

Carry this forward. An energy balance can rule out a design before you know its detailed construction.

Foundations of thermodynamics

A useful model starts with a boundary.

Full visual lesson ↗

01 · Calculate

Does a wider tank fill more slowly?

Water enters at 2 L/s and leaves at 1 L/s. A tank has area 2 m². Find its level rise after 60 s and its mass gain. Use ρ = 1,000 kg/m³.

A small hint

Volume gain is independent of area; height gain is not.

Explore the visual solution

Level rise after 60 s30 mm

Accumulated $volume = 60$ L; mass $gain = 60 \mathrm{kg}$ . The picture starts from a height of 0.40 m.

What if you change it?Tank area2 m²

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Net gain in 60 s = ( $2 - 1$ ) × $60 = 60$ $\mathrm{L} = 0.060$ m³.

Carry this forward. Geometry changes how accumulation appears, not how much mass enters.

02 · Reason

Choose the boundary for a pump.

You want the pressure rise produced by a pump. Compare following a fixed parcel of water with placing a fixed boundary around the pump inlet and outlet. Which is usually simpler for steady operation?

A small hint

Ask whether you want to follow material or measure a device continuously.

Explore the visual solution

DescriptionControl volume

Useful for steady device measurements. Shaft work enters through the device boundary.

What if you change it?

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

A material boundary follows the same water; by definition no mass crosses that boundary.

Carry this forward. A system choice is useful because it simplifies the question, not because one apparatus has only one valid boundary.

03 · Challenge

How long until the tank overflows?

A tank has area 0.50 m² and 0.20 m of freeboard. Inflow is 4 L/s; outflow is 1 L/s. Estimate time to overflow. Then explain why the model ends there.

A small hint

Available volume divided by net inflow gives the time.

Explore the visual solution

Time to reach the rim33.33 s

Remaining $capacity = 100$ L. Divide by 3 L/s; the picture shows the tank at the overflow threshold.

What if you change it?Tank area; 0.20 m freeboard0.5 m²

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Available $volume = 0.50 \times 0.20 = 0.100$ $\mathrm{m}^3 = 100$ L.

Carry this forward. A correct equation still has a domain of validity. Identify the physical event that ends it.

Foundations of thermodynamics

Closed is not the same as isolated.

Full visual lesson ↗

01 · Reason

Can insulation make a gas isolated?

A sealed, insulated piston-cylinder is compressed by an external force. No gas escapes. Classify the gas and identify the energy transfer.

A small hint

Insulation blocks heat, not a force acting through a moving boundary.

Explore the visual solution

Adiabatic closed-system energy changeΔU = +30 J

Q = 0; W = −30 J. ΔU = Q − W. The piston position is schematic.

What if you change it?Compression work input30 J

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

The system is closed because its mass is fixed.

Carry this forward. A moving boundary can transfer energy while remaining impermeable to mass.

02 · Calculate

A steady stream through a device.

A tank receives and releases 2 kg/s. How much mass crosses its inlet in 30 s? How much mass accumulates?

A small hint

Gross transfer through one port is not the same as net accumulation.

Explore the visual solution

Mass entering60 kg

Mass leaving60 kg

Mass accumulated0 kg

What if you change it?Elapsed time30 s

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Inlet $mass = 2 \times 30 = 60 \mathrm{kg}$ ; outlet mass is also 60 kg.

Carry this forward. Steady behavior permits continuous throughput.

03 · Challenge

Move the boundary around a heater.

A resistor immersed in water receives 200 J of electrical energy. Compare the water alone with water plus resistor. Neglect losses and assume the resistor ends at its initial internal energy.

A small hint

The same energy may cross different chosen boundaries through different mechanisms.

Explore the visual solution

Transfer across the chosen boundaryHeat

For the original process, water gains 200 J. Changing the boundary changes the transfer label, not the energy result.

What if you change it?

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

For water alone, energy crosses the resistor-water interface because of a temperature difference: heat transfer.

Carry this forward. Heat and work labels depend on the mechanism at the selected boundary.

Foundations of thermodynamics

What changes when you split a system?

Full visual lesson ↗

01 · Calculate

Scale the sample, keep the state.

A 2 kg sample occupies 0.002 m³ and has specific internal energy 80 kJ/kg. At the same state, find V, U, and density for 6 kg.

A small hint

Volume and total energy scale with mass; their mass-specific values remain fixed.

Explore the visual solution

Total internal energy480 kJ

Volume of the sample0.006 m³

$Density = 1,000 \mathrm{kg}$ /m³; $u = 80 \mathrm{kJ}$ /kg. Both stay constant.

What if you change it?Sample mass6 kg

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Specific volume is 0.002/ $2 = 0.001$ m³/kg.

Carry this forward. A property can be extensive even when its corresponding specific property is intensive.

02 · Calculate

Will these temperatures average to 50°C?

Mix 1 kg of water at 80°C with 3 kg at 20°C in an insulated container. Assume constant equal specific heats, no phase change, and negligible container heat capacity. Find the equilibrium temperature.

A small hint

Energy lost by the hotter water equals energy gained by the colder water.

Explore the visual solution

Equilibrium mixture temperature35°C

Tf = ( $1 \times 80 + 3 \times 20$ )/( $1 + 3$ ). Equal constant specific heats; no heat loss.

What if you change it?Cold-water mass3 kg

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Write 1c( $Tf - 80$ ) + 3c( $Tf - 20$ ) = 0.

Carry this forward. The warmer sample can have a higher temperature yet contain less mass and lower total energy than the colder sample.

03 · Challenge

Steady or equilibrium?

A wall has one face maintained at 80°C and the other at 20°C. The temperature profile no longer changes with time. Is the wall in thermal equilibrium?

A small hint

A time derivative can be zero while a spatial gradient is not.

Explore the visual solution

Thermal stateSteady, not equilibrium

Temperature is constant in time. The line shows its variation across wall position; the arrow shows heat-flow direction.

What if you change it?Hot-face temperature80 °C

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

The profile is steady because each location stays at a constant temperature over time.

Carry this forward. Steady means unchanging in time; equilibrium additionally excludes the driving forces for spontaneous change.

Foundations of thermodynamics

The zero on a gauge is a choice.

Full visual lesson ↗

01 · Calculate

Read a vacuum gauge.

A vessel has gauge $pressure -25 \mathrm{kPa}$ at an atmospheric pressure of 98 kPa. Find its absolute pressure.

A small hint

Add the local atmospheric reference, even when the gauge value is negative.

Explore the visual solution

Absolute vessel pressure73 kPa

-25 kPa $gauge + 98 \mathrm{kPa}$ atmosphere. Bars share the absolute-pressure reference.

What if you change it?Gauge pressure-25 kPa

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Use $p_{abs}=p_g+p_{atm}$ .

Carry this forward. An ordinary absolute-pressure measurement cannot be negative.

02 · Reason

Twice the Celsius reading?

A gas in a rigid sealed vessel warms from 20°C to 40°C. Treat it as ideal. Does its absolute pressure double?

A small hint

For fixed mass and volume, $\frac{p_2}{p_1}=\frac{T_2}{T_1}$ with temperatures in kelvin.

Explore the visual solution

Initial temperature293.15 K

Final temperature313.15 K

Final / initial absolute pressure1.0682 ×

The fractional pressure increase is 6.82%. Fixed mass, fixed volume, ideal gas.

What if you change it?Final temperature40 °C

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Convert: $T_1 = 293.15 \mathrm{K}$ and $T_2 = 313.15 \mathrm{K}$ .

Carry this forward. A 20°C change equals a 20 K change, but 40°C/20°C is not a valid gas-law temperature ratio.

03 · Challenge

Can a loaded piston stay in balance?

A frictionless horizontal piston has area 0.010 m². Gas on one side is at 250 kPa absolute and the atmosphere on the other is at 100 kPa. What additional opposing force holds it stationary?

A small hint

Atmospheric pressure already supplies part of the opposing force.

Explore the visual solution

Additional opposing force1,500 N

( $250 - 100$ ) $\mathrm{kPa} \times 1,000$ $\mathrm{Pa}/\mathrm{kPa} \times 0.010$ m².

What if you change it?Gas absolute pressure250 kPa

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Net pressure $difference = 250 - 100 = 150 \mathrm{kPa}$ .

Carry this forward. Use the pressure difference when calculating the net force across a surface.

Foundations of thermodynamics

Energy has more than one place to hide.

Full visual lesson ↗

01 · Calculate

Double the speed.

A 2 kg cart speeds up from 3 m/s to 6 m/s. Find the kinetic-energy increase, then compare final and initial kinetic energies.

A small hint

Subtract the two squared-speed energies, not the energy evaluated at the speed difference.

Explore the visual solution

Initial kinetic energy9 J

Final kinetic energy36 J

Kinetic-energy increase27 J

For m = 2 kg: ΔKE = ½ × 2 × (6² − 3²).

What if you change it?Final speed6 m/s

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Initial $KE = \frac12 \times 2 \times 3$ ² = 9 J.

Carry this forward. A difference of squares is not the square of a difference.

02 · Calculate

How much power does a lift need?

A lift raises 20 kg at a constant speed through 5 m in 10 s. Neglect losses. Find the energy increase and average input power.

A small hint

At constant speed, kinetic energy does not change.

Explore the visual solution

Gravitational energy increase981 J

Rise = 5 m. Average power over 10 s = 98.1 W. Drawing is schematic.

What if you change it?Vertical rise in 10 s5 m

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

$\Delta PE = 20 \times 9.81 \times 5 = 981 \mathrm{J}$ .

Carry this forward. A force can do work even when acceleration is zero.

03 · Challenge

Where does the falling weight’s energy go?

A 10 kg weight falls 2 m and drives a paddle in 1 kg of water. The weight starts and ends at rest. Assume all lost gravitational energy becomes water internal energy, with no heat loss. Use $c = 4,180 \mathrm{J}$ /(kg·K). Find the water temperature rise.

A small hint

Connect the mechanical-energy loss to mcΔT.

Explore the visual solution

Water temperature increase0.0469 K

Lost gravitational energy = 196.2 J. It enters the water as paddle work.

What if you change it?Weight drop2 m

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

The weight loses $10 \times 9.81 \times 2 = 196.2 \mathrm{J}$ of gravitational energy.

Carry this forward. A temperature increase does not prove that the energy entered as heat.