An object with a mass of $60 g$ is dropped into $700 mL$ of water at $0^@C$ . If the object cools by $40 ^@C$ 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 · 2017-07-16T09:44:54

The specific heat is $=29.3kJkg^-1K^-1$

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

For the cold water, $Delta T_w=24ºC$

For the object $DeltaT_o=40ºC$

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

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

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

Mass of the object is $m_0=0.060kg$

Mass of the water is $m_w=0.7kg$

$0.06*C_o*40=0.7*4.186*24$

$C_o=(0.7*4.186*24)/(0.06*40)$

$=29.3kJkg^-1K^-1$

An object with a mass of 60 g60 g is dropped into 700 mL700 mL of water at 0^@C0^@C. If the object cools by 40 ^@C40 ^@C and the water warms by 24 ^@C24 ^@C, what is the specific heat of the material that the object is made of?