An object with a mass of $30 g$ is dropped into $750 mL$ of water. If the object cools by $15 ^@C$ and the water warms by $15 ^@C$ , what is the specific heat of the material that the object is made of?

Question

1854 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-01T17:03:54

The specific heat is $=104.65kJkg^-1K^-1$

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water, $Delta T_w=15º$

For the object $DeltaT_o=15º$

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

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

$m_0 C_o*15 = m_w* 4.186 *15$

$0.030*C_o*15=0.75*4.186*15$

$C_o=(0.75*4.186*15)/(0.030*15)$

$=104.65kJkg^-1K^-1$

An object with a mass of 30 g30 g is dropped into 750 mL750 mL of water. If the object cools by 15 ^@C15 ^@C and the water warms by 15 ^@C15 ^@C, what is the specific heat of the material that the object is made of?