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?

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Answer 1 · 2016-08-13T09:21:30

Given
$m_o->"Mass of the object"=90g$

$v_w->"Volume of water object"=750mL$

$Deltat_o->"Rise of temperature of water"=4^@C$

$Deltat_w->"Fall of temperature of the object"=30^@C$

$d_w->"Density of water"=1g/(mL)$

$m_w->"Mass of water"$
$=v_wxxd_w=750mLxx1g/(mL)=750g$

$s_w->"Sp.heat of water"=1calg^"-1"""^@C^-1$

$"Let "s_o->"Sp.heat of the object"$

Now by calorimetric principle

Heat lost by object = Heat gained by water

$=>m_o xx s_o xxDeltat_o=m_wxxs_wxxDeltat_w$

$=>90xxs_o xx30=750xx1xx4$

$=>s_o=3000/2700=10/9$

$~~1.11calg^"-1"""^@C^-1$

An object with a mass of 90 g90 g is dropped into 750 mL750 mL of water at 0^@C0^@C. If the object cools by 30 ^@C30 ^@C and the water warms by 4 ^@C4 ^@C, what is the specific heat of the material that the object is made of?