# An object with a mass of 160 g is dropped into 880 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 5 ^@C, what is the specific heat of the material that the object is made of?

Feb 3, 2016

$1004.35 j k {g}^{- 1} {K}^{- 1}$

#### Explanation:

suppose,
the specific heat of the object is $S$
now,
we know,
$1 m l$ water= $1 g m$
$\therefore 880 m l$ water=$880 g m$
$= 0.88 k g$
and we know,
heat loss= heat gain
$0.16 S \times \left(120 - 5\right) = 0.88 \times 4200 \times 5$
$\implies 18.4 S = 18480$
$\implies S = 1004.35 j k {g}^{- 1} {K}^{- 1}$