Well-mixed tank heating/cooling model (repaired): Water enters a perfectly mixed tank at flow rate Q0 and temperature T0 and leaves at flow rate Q with exit temperature T. Heat losses are negligible. The tank volume V is constant (level control maintained). Which dynamic energy balance best describes the temperature change with time?

Introduction / Context:Dynamic temperature modeling of a continuously stirred tank under negligible heat loss is foundational in process control. It provides the transfer function from inlet temperature disturbances to outlet temperature and underpins controller design for heat exchangers and storage tanks.Given Data / Assumptions:

• Tank is perfectly mixed; exit temperature equals tank temperature T(t).
• Inlet flow rate and temperature: Q0, T0; outlet flow rate: Q.
• Negligible heat loss; liquid properties constant; overall heat capacity per unit volume normalized to 1 for clarity (or absorbed into V).
• Liquid holdup (volume) V is constant (level control present so dV/dt = 0).

Concept / Approach:The unsteady-state energy balance is: rate of energy accumulation = energy in − energy out. With constant density and heat capacity absorbed into units, this reduces to V * dT/dt = Q0 * T0 − Q * T. For many level-controlled cases Q0 = Q, which simplifies to V * dT/dt = Q0 (T0 − T). This is the canonical first-order model used to derive time constant τ = V/Q0.

Step-by-Step Solution:

Verification / Alternative check:Dimensional check: [Q0]*[T] has units of volume*temperature/time matching [V]*[dT/dt]. Solution form is first-order: dT/dt = (Q0/V) (T0 − T).

Why Other Options Are Wrong:

Common Pitfalls:Forgetting to include holdup (V) or assuming V varies with time. If properties are not normalized, include ρ*Cp explicitly.

Final Answer:Q0 (T0 - T) = V * dT/dt