An object with a mass of $60 g$ is dropped into $700 mL$ of water at $0^@C$ . If the object cools by $40 ^@C$ and the water warms by $16 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2016-02-17T18:09:27

$19553.33J//kg.K$

Conservation of thermal energy : Heat energy lost by water = heat energy gained by object.

$therefore (mcDeltaT)_(water) = (mcDeltaT)_(object)$

$therefore (0.700)(4190)(16)=(0.060)(c)(40)$

$therefore c=19553.33J//kg.K$

An object with a mass of 60 g60 g is dropped into 700 mL700 mL of water at 0^@C0^@C. If the object cools by 40 ^@C40 ^@C and the water warms by 16 ^@C16 ^@C, what is the specific heat of the material that the object is made of?