# Question f723a

Mar 11, 2016

Heat capacity of the sample  =1.44J//gm//"^oC, rounded to second place in decimal.
Heat capacity of pure gold is known to be =0.129J//gm//"^oC
The value found in the experiment appears exceptionally high.

#### Explanation:

As the quantities have been given in CGS units, therefore, problem has been worked out in same units. Final answer posted in the asked units.

Value used

Specific heat of water$= 1 c a l / g m = 4.19 k J / k g$

Let the specific heat of the sample $= {s}_{s}$

Amount of heat exchanged is given as $\Delta Q = m s \Delta t$,
where $m$ is the mass, $s$ is the specific heat and $\Delta t$ is change in temperature.

Heat gained by $15.5 g m$ of water to change from at ${24.4}^{o} C$ to water at ${27.0}^{o} C$,

$\Delta {Q}_{g a i n e d} = m s \Delta {t}_{w} = 15.5 \times 1 \times \left(27.0 - 24.4\right) = 40.3 c a l$

Simlarly heat lost by $2.58 g$ of sample is given as
$\Delta {Q}_{l o s t} = m {s}_{s} \Delta {t}_{s} = 2.58 \times {s}_{s} \times \left(72.3 - 27.0\right) = 116.874 {s}_{s}$

Since, $\Delta H e a {t}_{l o s t} = \Delta H e a {t}_{g a i n e d}$

$\therefore 116.874 {s}_{s} = 40.3$
$\implies {s}_{s} = \frac{40.3}{116.874}$
or ${s}_{s} = 0.34481 c a l / g m$
Changing to desired units. Recall $1 c a l = 4.184 J$
Heat capacity of the sample =1.44J//gm//"^oC#, rounded to second place in decimal.