# Determination of Specific Heat of Liquid

## Determination of Specific Heat of Liquid

If volume, radiating surface area, nature of surface, initial temperature and surrounding of water and given liquid are equal and they are allowed to cool down by radiation then rate of loss of heat and fall in temperature of both will be same.

i.e.,

$${{\left( \frac{dQ}{dt} \right)}_{Water}}={{\left( \frac{dQ}{dt} \right)}_{Liquid}}$$.

$$\left( {{m}_{W}}{{C}_{W}}+W \right)\left( \frac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{{{t}_{1}}} \right)=\left( {{m}_{l}}{{C}_{l}}+W \right)\left( \frac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{{{t}_{2}}} \right)$$.

$$\left( \frac{{{m}_{W}}{{C}_{W}}+W}{{{t}_{1}}} \right)=\left( \frac{{{m}_{l}}{{C}_{l}}+W}{{{t}_{2}}} \right)$$.

W = mcCc = Water equivalent of Calorimeter.

Where,

mc = Mass of Calorimeter,

Cc = Specific heat of Calorimeter.

If densities of water and liquid are ρ and θ’ respectively, then mW = VρW and ml = Vρl.

$$Specific\,\,heat\,\,of\,\,liquid\left( {{C}_{l}} \right)=\frac{1}{{{m}_{l}}}\left[ \frac{{{t}_{1}}}{{{t}_{W}}}\left( {{m}_{W}}{{c}_{W}}+W \right)-W \right]$$.