A certain time will be needed to warm up the iron to a certain temperature before ironing. The energy generated by the wires partly lost to the ambient by heat transfer. For a given ironing temperature, the warm up time and the quantity of heat loss need to be determined.

Energy balance of this warming up process:

      E_in - E_out = E_storage

During this process, the temperature of the base becomes higher than the ambient air temperature. So convection and radiation heat transfer exist.

      E_in = E_{generate by wires} η

      E_out= E_convection + E_radiation

1. If all the energy transferred to the base is stored to increase the temperature of the base, the time will be shortest. In another word, assuming no heat loss to the ambient by convection and radiation.

      E_in - E_out = E_storage

      E_{storage =} E_in - E_out

The heat generated by the wires is:

Energy stored in the base is:

      E_storage = c_pm (T_final - T_initial)

      m = ρAδ

Plug the numbers, the solution for time t is:

      t = 342 s

2. Consider the convection and radiation heat loss from the iron after the temperature reaches 140 ^oC. Because temperature keeps at 140^o C, no energy is stored.
  
Comparing the heat loss and heat input.

      98.8/100 = 0.98 = 98.8%

The power of this iron is not enough to keep the temperature at 140 ^oC during ironing because the clothes will absorb more heat than the ambient.