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_{p}m (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 ^{o}C. 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
^{o}C during ironing because the clothes will absorb more heat
than the ambient.
