# Question dc2ff

Feb 4, 2018

Consider,

$q = m {C}_{s} \Delta T$

Hence,

8.75*10^5J = 2775.0g * C_s * 25°C
therefore C_s approx (12.6J)/(g*°C)#

is the specific heat of this "specific" piece of wood.

Given its distribution of lignin and other structural polysaccharides, along with the specific content of its cells, it is fairly unique. So if we're doing more precise work we would need to continually experimentally evaluate wood specific heat.

On the other hand, if it was a pure element (e.g. silver or copper) we could reliably consider its specific heat in calculations involving enthalpy.