An object with a mass of $24 g$ is dropped into $6400 mL$ 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 · 2018-01-25T07:05:46

The specific heat is $=55.8 kJkg^-1K^-1$

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

For the cold water, $Delta T_w=3ºC$

For the object $DeltaT_o=60ºC$

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

The specific heat of water $C_w=4.186kJkg^-1K^-1$

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

The mass of the object is $m_o=0.024kg$

The mass of the water is $m_w=6.4kg$

$0.024*C_o*60=6.4*4.186*3$

$C_o=(6.4*4.186*3)/(0.024*60)$

$=55.8 kJkg^-1K^-1$

An object with a mass of 24 g24 g is dropped into 6400 mL6400 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?