# An object with a mass of 32 g is dropped into 250 mL of water at 0^@C. If the object cools by 60 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

Aug 18, 2016

Given
${m}_{o} \to \text{Mass of the object} = 32 g$

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

$\Delta {t}_{w} \to \text{Rise of temperature of water} = {3}^{\circ} C$

$\Delta {t}_{o} \to \text{Fall of temperature of the object} = {60}^{\circ} C$

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

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

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

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

Now by calorimetric principle

Heat lost by object = Heat gained by water

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

$\implies 32 \times {s}_{o} \times 60 = 250 \times 1 \times 3$

$\implies {s}_{o} = \frac{250 \times 3}{32 \times 60}$

$\approx 0.39 c a l {g}^{\text{-1}} ^ \circ {C}^{-} 1$

Aug 18, 2016

masss of water= 250/1000= 1/4 kg
specific heat capacity of water= 42000J/$k {g}^{0} C$
Change in temperature= ${3}^{0} C$
Heat gained by water= mst = 1/442003= 3150
this is the heat lost by object
mass of object =32g =.032 kg
Change in temperature= ${60}^{0} C$
Specific heat capacity of object = H/mt
=3150/.032*60
=1640.625 J / $k {g}^{0} C$