MechanicalThermodynamicseasyFE ExamCoursework

Heat Added to a Rigid Tank

A rigid, sealed tank contains $m = 2.0\ \text{kg}$ of air. The air is heated from $300\ \text{K}$ to $500\ \text{K}$ .

For air, $c_v = 0.718\ \text{kJ/kg·K}$ .

Find the heat added to the air.

Given

m	2 kg	Mass of air
T1	300 K	Initial temperature
T2	500 K	Final temperature
cv	0.718 kJ/kg*K	Specific heat at constant volume

Hint 1

Start from the first law for a closed system: Q − W = ΔU.

Hint 2

What does 'rigid' tell you about the work term?

Hint 3

With W = 0, Q = ΔU = mc_vΔT. Note it is c_v, not c_p.

Worked solution — try the problem first

First law for a closed system:

$Q - W = \Delta U$

The tank is rigid, so the volume cannot change and no boundary work is done:

$W = \int P\,dV = 0$

Therefore $Q = \Delta U$ .

Internal energy change for an ideal gas with constant specific heat uses $c_v$ — the constant-volume specific heat — regardless of the process. Here the process happens to be constant volume too, but that is not why $c_v$ appears:

$\Delta U = m c_v (T_2 - T_1) = 2.0 \times 0.718 \times (500 - 300)$

$= 2.0 \times 0.718 \times 200 = 287.2\ \text{kJ}$

$\boxed{Q = 287.2\ \text{kJ}}$

Concepts:First law of thermodynamicsClosed systemsSpecific heats

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