An object with a mass of $4 g$ is dropped into $650 mL$ of water at $0^@C$ . If the object cools by $300 ^@C$ and the water warms by $30 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-02-18T09:42:13

The specific heat of the material is $=68.0 KJkg^-1°C^-1$

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water, $Delta T_w=30ºC$

For the object $DeltaT_o=300ºC$

$M_o C_o (DeltaT_o) = Mw C_w (Delta_w)$

$C_w=4.186KJkg^-1ºC^-1$

$0,004* C_o*300 = 0,6500* 4.186 *30$

$C_o = (0.650*4.186*30)/(0.004*300) = 68.0 KJkg^-1°C^-1$

An object with a mass of 4 g4 g is dropped into 650 mL650 mL of water at 0^@C0^@C. If the object cools by 300 ^@C300 ^@C and the water warms by 30 ^@C30 ^@C, what is the specific heat of the material that the object is made of?