# An object with a mass of 90 g is dropped into 750 mL of water at 0^@C. If the object cools by 30 ^@C and the water warms by 4 ^@C, what is the specific heat of the material that the object is made of?

Aug 13, 2016

Given
${m}_{o} \to \text{Mass of the object} = 90 g$

${v}_{w} \to \text{Volume of water object} = 750 m L$

$\Delta {t}_{o} \to \text{Rise of temperature of water} = {4}^{\circ} C$

$\Delta {t}_{w} \to \text{Fall of temperature of the object} = {30}^{\circ} C$

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

${m}_{w} \to \text{Mass of water}$
$= {v}_{w} \times {d}_{w} = 750 m L \times 1 \frac{g}{m L} = 750 g$

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

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

Now by calorimetric principle

Heat lost by object = Heat gained by water

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

$\implies 90 \times {s}_{o} \times 30 = 750 \times 1 \times 4$

$\implies {s}_{o} = \frac{3000}{2700} = \frac{10}{9}$

$\approx 1.11 c a l {g}^{\text{-1}} ^ \circ {C}^{-} 1$