riaology
Unit 02 · Lesson 06 of 09

How much energy changes the temperature?

A heat capacity is a slope you can integrate.

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

What you’ll be able to do
  • Define the slopes.
  • Derive the ideal-gas difference.
  • Integrate before approximating.
Before you touch the controls

Will a constant heat-capacity approximation always match a variable-property calculation?

An experiment you can change

Integrate a temperature-dependent specific heat

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

Piston cylinder, volume 2 liters; 20 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveIntegrate a temperature-dependent specific heat. Temperature (K) versus Specific heat (kJ/(kg K)). Selected result 223.5 kJ/kg. At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.03000.384500.76600Specific heat (kJ/(kg K))Temperature (K)
Δu=0.7(600−300)+0.00012(6002−3002)=223.5 kJ/kg\Delta u=0.7(600-300)+\frac{0.0001}{2}(600^2-300^2)=223.5\,\mathrm{kJ/kg}

At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.

Model assumptions and units

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.0001T kJ/(kg K), R=0.287 kJ/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Δu=∫T1T2cv(T) dT,Δh=∫T1T2cp(T) dT,cp−cv=R\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT,\quad \Delta h=\int_{T_1}^{T_2}c_p(T)\,dT,\quad c_p-c_v=R
01

Define the slopes.

Specific heats are partial derivatives of internal energy and enthalpy at specified constraints. For an ideal gas u and h depend only on temperature, so these derivatives determine their changes along any process.

cv=(∂u∂T)v,cp=(∂h∂T)pc_v=\left(\frac{\partial u}{\partial T}\right)_v,\quad c_p=\left(\frac{\partial h}{\partial T}\right)_p
02

Derive the ideal-gas difference.

Specific enthalpy is u+pvu+pv. With pv=RTpv=RT, h=u+RTh=u+RT. Differentiating with respect to temperature gives cp=cv+Rc_p=c_v+R. Use the same mass or molar basis for all three quantities.

h=u+RT⟹cp−cv=Rh=u+RT\quad\Longrightarrow\quad c_p-c_v=R
03

Integrate before approximating.

The area beneath cv(T)c_v(T) gives the specific internal-energy change; the area beneath cp(T)c_p(T) gives the enthalpy change. A constant c is an approximation, often selected at a representative temperature.

Δu=∫T1T2cv(T) dT≈cv,avg(T2−T1)\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT\approx c_{v,avg}(T_2-T_1)
04

Liquids and solids have another approximation.

For incompressible substances, specific volume is approximately constant and cp≈cv≈cc_p\approx c_v\approx c. Internal-energy change follows the heat-capacity integral. Enthalpy also contains the pressure contribution; neglect it only when small for the question.

Δh≈∫T1T2c(T) dT+v(p2−p1)\Delta h\approx\int_{T_1}^{T_2}c(T)\,dT+v(p_2-p_1)
A worked example

A variable-cv teaching model.

  1. Let cv(T)=0.70+0.0001Tc_v(T)=0.70+0.0001T in kJ/(kg K), with T1=300 KT_1=300\,\mathrm K and T2=600 KT_2=600\,\mathrm K.
  2. Integrate: Δu=0.70(600−300)+0.00005(6002−3002)=223.5 kJ/kg\Delta u=0.70(600-300)+0.00005(600^2-300^2)=223.5\,\mathrm{kJ/kg}.
  3. For R=0.287 kJ/(kg K)R=0.287\,\mathrm{kJ/(kg\,K)}, Δh=223.5+0.287(300)=309.6 kJ/kg\Delta h=223.5+0.287(300)=309.6\,\mathrm{kJ/kg}.
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

Integrate internal energy

For the teaching gas cv(T) = 0.70+0.00010.70 + 0.0001T, heat from 300 to 600 K. Find specific internal-energy change.

Hint

Integrate the constant and linear terms separately.

Explore the visual solution
Piston cylinder, volume 2 liters; 20 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveIntegrate a temperature-dependent specific heat. Temperature (K) versus Specific heat (kJ/(kg K)). Selected result 223.5 kJ/kg. At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.03000.384500.76600Specific heat (kJ/(kg K))Temperature (K)
Δu=0.7(600−300)+0.00012(6002−3002)=223.5 kJ/kg\Delta u=0.7(600-300)+\frac{0.0001}{2}(600^2-300^2)=223.5\,\mathrm{kJ/kg}

At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.

Model assumptions and units

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.0001T kJ/(kg K), R=0.287 kJ/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Δu=∫T1T2cv(T) dT,Δh=∫T1T2cp(T) dT,cp−cv=R\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT,\quad \Delta h=\int_{T_1}^{T_2}c_p(T)\,dT,\quad c_p-c_v=R

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

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.0001T kJ/(kg K), R=0.287 kJ/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Δu=∫T1T2cv(T) dT,Δh=∫T1T2cp(T) dT,cp−cv=R\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT,\quad \Delta h=\int_{T_1}^{T_2}c_p(T)\,dT,\quad c_p-c_v=R
2 · Compare

Constant-pressure heat

For the same teaching gas with R=0.287kJR = 0.287 \mathrm{kJ}/(kg K), heat from 300 to 600 K at constant pressure with only boundary work. Find heat per kilogram.

Hint

Use cp = cv + R and the enthalpy change.

Explore the visual solution
Piston cylinder, volume 2 liters; 20 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveIntegrate a temperature-dependent specific heat. Temperature (K) versus Specific heat (kJ/(kg K)). Selected result 309.6 kJ/kg. At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.03000.5234501.047600Specific heat (kJ/(kg K))Temperature (K)
Δh=0.9869999999999999(600−300)+0.00012(6002−3002)=309.6 kJ/kg\Delta h=0.9869999999999999(600-300)+\frac{0.0001}{2}(600^2-300^2)=309.6\,\mathrm{kJ/kg}

At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.

Model assumptions and units

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.0001T kJ/(kg K), R=0.287 kJ/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Δu=∫T1T2cv(T) dT,Δh=∫T1T2cp(T) dT,cp−cv=R\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT,\quad \Delta h=\int_{T_1}^{T_2}c_p(T)\,dT,\quad c_p-c_v=R

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

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.0001T kJ/(kg K), R=0.287 kJ/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Δu=∫T1T2cv(T) dT,Δh=∫T1T2cp(T) dT,cp−cv=R\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT,\quad \Delta h=\int_{T_1}^{T_2}c_p(T)\,dT,\quad c_p-c_v=R
3 · Challenge the model

A wider temperature interval

Heat the teaching gas from 300 to 900 K at constant volume. Find internal-energy change and compare with cv at 300 K multiplied by the interval.

Hint

The approximation uses 0.73 × 600; the integral includes the increasing slope.

Explore the visual solution
Piston cylinder, volume 2 liters; 20 joules of heat in and 0 joules of work by gasFixed mass; boundary may moveIntegrate a temperature-dependent specific heat. Temperature (K) versus Specific heat (kJ/(kg K)). Selected result 456 kJ/kg. At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.03000.3956000.79900Specific heat (kJ/(kg K))Temperature (K)
Δu=0.7(900−300)+0.00012(9002−3002)=456 kJ/kg\Delta u=0.7(900-300)+\frac{0.0001}{2}(900^2-300^2)=456\,\mathrm{kJ/kg}

At constant volume with no other work Q = ΔU. At constant pressure with only boundary work Q = ΔH. Integrating avoids silently assuming constant specific heat.

Model assumptions and units

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.0001T kJ/(kg K), R=0.287 kJ/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Δu=∫T1T2cv(T) dT,Δh=∫T1T2cp(T) dT,cp−cv=R\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT,\quad \Delta h=\int_{T_1}^{T_2}c_p(T)\,dT,\quad c_p-c_v=R

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

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.0001T kJ/(kg K), R=0.287 kJ/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Δu=∫T1T2cv(T) dT,Δh=∫T1T2cp(T) dT,cp−cv=R\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT,\quad \Delta h=\int_{T_1}^{T_2}c_p(T)\,dT,\quad c_p-c_v=R
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