An object with a mass of $32 g$ is dropped into $250 mL$ of water at $0^@C$ . If the object cools by $16 ^@C$ and the water warms by $3 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2016-08-11T08:59:41

Given
$m_o->"Mass of the object"=32g$

$v_w->"Volume of water object"=250mL$

$Deltat_w->"Rise of temperature of water"=3^@C$

$Deltat_o->"Fall of temperature of the object"=16^@C$

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

$m_w->"Mass of water"$
$=v_wxxd_w=250mLxx1g/(mL)=250g$

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

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

Now by calorimetric principle

Heat lost by object = Heat gained by water

$=>m_o xx s_o xxDeltat_o=m_wxxs_wxxDeltat_w$

$=>32xxs_o xx16=250xx1xx3$

$=>s_o=(250xx3)/(32xx16)$

$~~1.46calg^"-1"""^@C^-1$

An object with a mass of 32 g32 g is dropped into 250 mL250 mL of water at 0^@C0^@C. If the object cools by 16 ^@C16 ^@C and the water warms by 3 ^@C3 ^@C, what is the specific heat of the material that the object is made of?