If the efficiency of an electric furnace is 96%, then 96% of the input energy is transformed into thermal energy. what is the other 4% of the electricity transformed into?

The 96% of input energy is obtained as work and remaining 4% of the electricity is also transformed into the heat, but this heat energy is utilized in the equipments of the furnace.

Given data:

The efficiency of the electric furnace is, .

Percentage of input energy transformed is, .

The efficiency of a system is defined as the ratio of input energy to the work output from the system. The mathematical expression for the efficiency is,

Here, W is the work output.

If the efficiency of an electric furnace is 96%, then 96% of the input energy shows up at a place in the form of thermal energy to get transferred to an airflow that is then pumped through some ducts and distributed around the house to heat the rooms. The other 4% is also converted to heat, but not in a place or form  that can go to heat the rooms in the house.  It just heats the furnace  housing, the blower motor, the air in the basement that's around the  furnace, etc.

Thus, we can conclude that the remaining 4% of the electricity is also transformed into the heat, but this heat energy is utilized in the equipments of the furnace.

Learn more about the mechanical efficiency here:

link

The easy answer is "heat". 
In this particular case, the easy answer isn't helpful.

I think we have to be a little more careful about exactly how
we define 'efficiency'.    I found this definition online:

       "the ratio of the useful work performed by a machine
          to the total energy expended or taken in"

I think the key word in this definition is "useful".

If the efficiency of an electric furnace is 96%, then 96% of the
input energy shows up at a place and in a form that it's available,
to get transferred to an airflow that is then pumped through some
ducts and distributed around the house to heat the rooms.

The other 4% is also converted to heat, but not in a place or form
that can go to heat the rooms in the house.  It just heats the furnace
housing, the blower motor, the air in the basement that's around the
furnace, etc.   That much of the input doesn't come out in a 'useful' form.


Here's another way I like to talk about 'efficiency', and I think it's particularly
relevant to the concept of 'useful':

Take a light bulb ... a good old-fashioned 60-watt incandescent light bulb.
It's designed to be used to light up a room, but it has other uses.  If you
have baby hamsters, guinea pigs, ferrets, possums, etc. ... any small
burrowing animal ... you can keep their beds nice and warm on a cold
night by hanging an incandescent light bulb under their box, because
an incandescent light bulb turns quite a bit of its input energy into heat. 
In fact, strange as it may seem, only about 10% of the energy that goes
into an incandescent bulb gets turned into light; the other 90% turns into
heat.

So here's something to think about:

-- IF you use an incandescent bulb for the purpose of adding light
to a room, then the bulb is 10% efficient.  The other 90% of the energy
you put into it comes out of the bulb in the form of heat, which isn't useful.

-- IF you use the same incandescent bulb for the purpose of heating
your hamster's box on a cold night, then the bulb is 90% efficient. 
The other 10% of the energy you put into it comes out of the bulb
in the form of light, which isn't useful.

The SAME device can have different efficiencies !  It depends on
what YOU use ir for, and which part of the output is useful to you.

Ibelieve the correct answer would be c.