An object with a mass of $120 g$ is dropped into $640 mL$ of water at $0^@C$ . If the object cools by $6 ^@C$ and the water warms by $24 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-11-23T13:54:54

The specific heat is $=89.3kJkg^-1K^-1$

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

For the cold water, $Delta T_w=24ºC$

For the object $DeltaT_o=6ºC$

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

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

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

The mass of the object is $m=0.120kg$

$0.120*C_o*6=0.640*4.186*24$

$C_o=(0.640*4.186*24)/(0.120*6)$

$=89.3kJkg^-1K^-1$

An object with a mass of 120 g120 g is dropped into 640 mL640 mL of water at 0^@C0^@C. If the object cools by 6 ^@C6 ^@C and the water warms by 24 ^@C24 ^@C, what is the specific heat of the material that the object is made of?