riaology
Unit 02 · Lesson 04 of 09

How does a closed system store energy?

A rigid wall stops boundary work, not every energy transfer.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do
  • Start with a material system.
  • Name each work mechanism.
  • Connect storage to state.
Before you touch the controls

Can an insulated rigid tank warm up?

An experiment you can change

A sealed gas receives heat and shaft work

Predict the direction of change, move one control, then explain the response using the equation.

Piston cylinder, volume 2 liters; 50 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveA sealed gas receives heat and shaft work. Heat input (kJ) versus Temperature change (K). Selected result 48.7465 K. Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.-13.928-4034.8193083.565100Temperature change (K)Heat input (kJ)
ΔT=50+202(0.718)=48.747 K\Delta T=\frac{50+20}{2(0.718)}=48.747\,\mathrm K

Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.

Model assumptions and units

Rigid sealed tank with 2 kg ideal gas, cv = 0.718 kJ/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb = 0, not zero shaft work.

ΔU=Q−Wb−Ws,ΔT=Q+Ws,inmcv\Delta U=Q-W_b-W_s,\qquad \Delta T=\frac{Q+W_{s,\mathrm{in}}}{mc_v}
01

Start with a material system.

No mass crosses a closed-system boundary. Heat and several types of work can still cross it. Keep kinetic and potential energy changes until the physical situation lets you neglect them.

ΔU+ΔKE+ΔPE=Q−W\Delta U+\Delta KE+\Delta PE=Q-W
02

Name each work mechanism.

With work by the system positive, expansion work is positive and shaft or electrical input is negative. A rigid tank has zero boundary work but may still have work input through a shaft or wires.

W=Wb+Ws+WeW=W_b+W_s+W_e
03

Connect storage to state.

For an ideal gas with constant cv, internal-energy change depends on temperature change. Density and pressure may change too, but they are not added as separate energy stores. For a real fluid, evaluate u from property data.

ΔU=mcv(T2−T1)\Delta U=m c_v(T_2-T_1)
04

Check the outcome.

An insulated rigid tank with positive shaft input increases internal energy. For the ideal-gas model temperature rises. Insulation specifies the absence of heat transfer; it does not specify constant temperature.

Q=0,Wb=0⟹ΔU=Ws,inQ=0,\quad W_b=0\quad\Longrightarrow\quad\Delta U=W_{s,in}
A worked example

Heat input and stirring in a rigid tank.

  1. Use m=2 kgm=2\,\mathrm{kg}, cv=0.718 kJ/(kg K)c_v=0.718\,\mathrm{kJ/(kg\,K)}, heat input 50 kJ, and shaft input 20 kJ.
  2. The signed work is Ws=−20 kJW_s=-20\,\mathrm{kJ}, so ΔU=50−(−20)=70 kJ\Delta U=50-(-20)=70\,\mathrm{kJ}.
  3. Then ΔT=70/[2(0.718)]=48.75 K\Delta T=70/[2(0.718)]=48.75\,\mathrm K.
Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Heat plus stirring

A rigid tank has 2 kg ideal gas with cv=0.718kJcv = 0.718 \mathrm{kJ}/(kg K). Heat input is 50 kJ and shaft input is 20 kJ. Find temperature change.

Hint

Work by the gas is −20 kJ, so subtracting work adds the input.

Explore the visual solution
Piston cylinder, volume 2 liters; 50 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveA sealed gas receives heat and shaft work. Heat input (kJ) versus Temperature change (K). Selected result 48.7465 K. Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.-13.928-4034.8193083.565100Temperature change (K)Heat input (kJ)
ΔT=50+202(0.718)=48.747 K\Delta T=\frac{50+20}{2(0.718)}=48.747\,\mathrm K

Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.

Model assumptions and units

Rigid sealed tank with 2 kg ideal gas, cv = 0.718 kJ/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb = 0, not zero shaft work.

ΔU=Q−Wb−Ws,ΔT=Q+Ws,inmcv\Delta U=Q-W_b-W_s,\qquad \Delta T=\frac{Q+W_{s,\mathrm{in}}}{mc_v}

Controls explore variations. The steps below retain the original question’s values.

Rigid sealed tank with 2 kg ideal gas, cv = 0.718 kJ/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb = 0, not zero shaft work.

ΔU=Q−Wb−Ws,ΔT=Q+Ws,inmcv\Delta U=Q-W_b-W_s,\qquad \Delta T=\frac{Q+W_{s,\mathrm{in}}}{mc_v}
2 · Compare

Cooling while stirring

The same tank loses 40 kJ of heat and receives 20 kJ of shaft work. Find temperature change.

Hint

Heat is signed; compare its loss with the work input.

Explore the visual solution
Piston cylinder, volume 2 liters; -40 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveA sealed gas receives heat and shaft work. Heat input (kJ) versus Temperature change (K). Selected result -13.9276 K. Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.-13.928-4034.8193083.565100Temperature change (K)Heat input (kJ)
ΔT=−40+202(0.718)=−13.928 K\Delta T=\frac{-40+20}{2(0.718)}=-13.928\,\mathrm K

Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.

Model assumptions and units

Rigid sealed tank with 2 kg ideal gas, cv = 0.718 kJ/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb = 0, not zero shaft work.

ΔU=Q−Wb−Ws,ΔT=Q+Ws,inmcv\Delta U=Q-W_b-W_s,\qquad \Delta T=\frac{Q+W_{s,\mathrm{in}}}{mc_v}

Controls explore variations. The steps below retain the original question’s values.

Rigid sealed tank with 2 kg ideal gas, cv = 0.718 kJ/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb = 0, not zero shaft work.

ΔU=Q−Wb−Ws,ΔT=Q+Ws,inmcv\Delta U=Q-W_b-W_s,\qquad \Delta T=\frac{Q+W_{s,\mathrm{in}}}{mc_v}
3 · Challenge the model

Offset the cooling exactly

The same tank loses 40 kJ of heat and receives 40 kJ shaft work. Find temperature change and explain whether the tank is isolated.

Hint

Equal opposing transfers can leave storage unchanged.

Explore the visual solution
Piston cylinder, volume 2 liters; -40 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveA sealed gas receives heat and shaft work. Heat input (kJ) versus Temperature change (K). Selected result 0 K. Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.0-4048.7473097.493100Temperature change (K)Heat input (kJ)
ΔT=−40+402(0.718)=0 K\Delta T=\frac{-40+40}{2(0.718)}=0\,\mathrm K

Work input is negative work by the gas. Heat can leave while shaft work still raises internal energy.

Model assumptions and units

Rigid sealed tank with 2 kg ideal gas, cv = 0.718 kJ/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb = 0, not zero shaft work.

ΔU=Q−Wb−Ws,ΔT=Q+Ws,inmcv\Delta U=Q-W_b-W_s,\qquad \Delta T=\frac{Q+W_{s,\mathrm{in}}}{mc_v}

Controls explore variations. The steps below retain the original question’s values.

Rigid sealed tank with 2 kg ideal gas, cv = 0.718 kJ/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb = 0, not zero shaft work.

ΔU=Q−Wb−Ws,ΔT=Q+Ws,inmcv\Delta U=Q-W_b-W_s,\qquad \Delta T=\frac{Q+W_{s,\mathrm{in}}}{mc_v}
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