An object with a mass of $150 g$ is dropped into $900 mL$ of water at $0^@C$ . If the object cools by $40 ^@C$ and the water warms by $16 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2018-05-14T06:35:10

The specific heat is $=10.05 kJkg^-1K^-1$

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

For the cold water, $Delta T_w=16ºC$

For the hot object $DeltaT_o=40ºC$

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

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 $m_o=0.150kg$

The mass of the water is $m_w=0.900kg$

$0.15*C_o*40=0.9*4.186*16$

$C_0=(0.9*4.186*16)/(0.15*40)$

$=10.05 kJkg^-1K^-1$

An object with a mass of 150 g150 g is dropped into 900 mL900 mL of water at 0^@C0^@C. If the object cools by 40 ^@C40 ^@C and the water warms by 16 ^@C16 ^@C, what is the specific heat of the material that the object is made of?