An object with a mass of $64 g$ is dropped into $560 mL$ of water at $0^@C$ . If the object cools by $80 ^@C$ and the water warms by $6 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2016-02-26T18:08:24

$2749.69J//kg.K$

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

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

$therefore 0.064xxcxx80=0.560xx4190xx6$

$therefore c=2749.69J//kg.K$

An object with a mass of 64 g64 g is dropped into 560 mL560 mL of water at 0^@C0^@C. If the object cools by 80 ^@C80 ^@C and the water warms by 6 ^@C6 ^@C, what is the specific heat of the material that the object is made of?