# An object with a mass of 15 g is dropped into 250 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 2 ^@C, what is the specific heat of the material that the object is made of?

Aug 11, 2016

$\approx 0.28 c a l {g}^{\text{-1}} ^ \circ {C}^{-} 1$

#### Explanation:

Given
${m}_{o} \to \text{Mass of the object} = 15 g$

${v}_{w} \to \text{Volume of water object} = 250 m L$

$\Delta {t}_{w} \to \text{Rise of temperature of water} = {2}^{\circ} C$

$\Delta {t}_{o} \to \text{Fall of temperature of the object} = {120}^{\circ} C$

${d}_{w} \to \text{Density of water} = 1 \frac{g}{m L}$

${m}_{w} \to \text{Mass of water}$
$= {v}_{w} \times {d}_{w} = 250 m L \times 1 \frac{g}{m L} = 250 g$

${s}_{w} \to {\text{Sp.heat of water"=1calg^"-1}}^{\circ} {C}^{-} 1$

$\text{Let "s_o->"Sp.heat of the object}$

Now by calorimetric principle

Heat lost by object = Heat gained by water

$\implies {m}_{o} \times {s}_{o} \times \Delta {t}_{o} = {m}_{w} \times {s}_{w} \times \Delta {t}_{w}$

$\implies 15 \times {s}_{o} \times 120 = 250 \times 1 \times 2$

$\implies {s}_{o} = \frac{250 \times 2}{15 \times 120} = \frac{5}{18}$

$\approx 0.28 c a l {g}^{\text{-1}} ^ \circ {C}^{-} 1$