An object with a mass of $10 g$ is dropped into $120 mL$ of water at $0^@C$ . If the object cools by $60 ^@C$ 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 · 2017-06-02T17:51:39

The specific heat is $=2.51kJkg^-1K^-1$

The heat is transferred from the hot object to the cold water.

For the cold water, $Delta T_w=3ºC$

For the object $DeltaT_o=60ºC$

$m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)$

$C_w=4.186kJkg^-1K^-1$

Let, $C_o$ be the specific heat of the object

$0.010*C_o*60=0.12*4.186*3$

$C_o=(0.12*4.186*3)/(0.010*60)$

$=2.51kJkg^-1K^-1$

An object with a mass of 10 g10 g is dropped into 120 mL120 mL of water at 0^@C0^@C. If the object cools by 60 ^@C60 ^@C and the water warms by 3 ^@C3 ^@C, what is the specific heat of the material that the object is made of?