about specific heat capacity

魔龍帝齊格益力多's answer is wrong (Sorry about that)

He said:
delta t=T
c=E/(mT)
________________
If heat is lost to surrounding,
E will be decreased
From the above formula,c is proportional to E.
The specific heat capacity is decreased.
_____________________________

In the experiment we done in school lab,the common reason that causes the error is energy losing to surrounding.We usually calculate the specific heat capacity which is lower than the standard value.
--------------------------------------------

But people always mess up with the value of E.

Actually, imagine, when you heat a water (same as metal), in ideal case, all energy get into water and surely the temperature of water will rise quickily as ALL the energy are absorbed by water.

But, in reality, the energy is used to heat up water AND surroundings (such as air, beaker, etc), so you need to have a larger energy output for heating up water by 1 degree Celcius, so E in experiment will be larger than that of ideal case.

( c is directly proportional to E )
so for experiment, E will be usally larger than the actual energy needed, so the calculated c will be larger.

I know that,sorry for my wrong answer..

The heat transfer equation is:

Q = mcT + L
where Q is the heat supplied
L is the heat lost to the surroundings
m is the mass of object invloved
c is the specific heat capacity of the object
T is the change of temperature

hence, c = (Q - L)/mT
If you neglect the heat lost to the surroundings in an experiment, then
c(experiment) = Q/mT

If heat loss is included, then
c (standard) = (Q - L)/mT
Clearly, c(experiment) > c(standard)
The experimental value is higher than the standard value.