An object with a mass of $45 g$ is dropped into $350 mL$ of water at $0^@C$ . If the object cools by $24 ^@C$ and the water warms by $8 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-06-21T11:42:12

The specific heat of the material is $=10.85kJkg^-1K^-1$

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

For the cold water, $Delta T_w=8ºC$

For the object $DeltaT_o=24ºC$

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

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

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

Mass of the object is $m_0=0.045kg$

Mass of the water is $m_w=0.35kg$

$0.045*C_o*24=0.35*4.186*8$

$C_o=(0.35*4.186*8)/(0.045*24)$

$=10.85kJkg^-1K^-1$

An object with a mass of 45 g45 g is dropped into 350 mL350 mL of water at 0^@C0^@C. If the object cools by 24 ^@C24 ^@C and the water warms by 8 ^@C8 ^@C, what is the specific heat of the material that the object is made of?