Answer
$Q_{dissipated}=2.4kW$
It is not necesary to turn the heater on when the motor runs at full load.
Work Step by Step
The $0.88$ of efficiency represents the fraction of energy that becomes to mechanical work at the rotor, the remaining part represents the one that becomes to thermal energy, then the dissipated heat is:
$Q_{dissipated}=20kW*(1-0.88)=2.4kW$
As you can see this dissipated heat is greater than the $2kW$ resistance heater, so it is not necesary to turn the heater on when the motor runs at full load.
