An object with a mass of 160 g is dropped into 880 mL of water at 0^@C. If the object cools by 18 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

Feb 8, 2016

${s}_{o b j e c t} = 2. \dot{4}$ $c a l / g m$

Explanation:

Assumption : Specific gravity of water at ${0}^{\circ} C$ is 1. (actually 0.9998)
$\implies$ weight of $880 m l$ of water is $880 g m$.

Now by Law of conservation of energy
Heat lost by object = Heat gained by water.

Also Heat gained or lost $= m \cdot s \cdot \Delta t$
where $m$ is the mass of object, $s$ its specific heat and$\Delta t$ Change in the temperature of the object.

Heat lost by object$= 160 \cdot {s}_{o b j e c t} \cdot 18$
Heat gained by water $= 880 \cdot 1 \cdot 8$
(Specific heat of water in cgs units is 1 calorie/gm)
Equating two we obtain
$160 \cdot {s}_{o b j e c t} \cdot 18 = 880 \cdot 1 \cdot 8$
${s}_{o b j e c t} = \frac{880 \cdot 8}{160 \cdot 18}$
${s}_{o b j e c t} = 2. \dot{4}$ $c a l / g m$