An object with a mass of $10 g$ is dropped into $240 mL$ of water at $0^@C$ . If the object cools by $120 ^@C$ and the water warms by $5 ^@C$ , what is the specific heat of the material that the object is made of?

Question

1511 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-13T07:38:23

The specific heat is $=4.186 kJkg^-1K^-1$

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

For the cold water, $Delta T_w=5ºC$

For the object $DeltaT_o=120ºC$

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

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 $m_o=0.01kg$

The mass of the water is $m_w=0.24kg$

$0.01*C_o*120=0.24*4.186*5$

$C_o=(0.24*4.186*5)/(0.01*120)$

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

An object with a mass of 10 g10 g is dropped into 240 mL240 mL of water at 0^@C0^@C. If the object cools by 120 ^@C120 ^@C and the water warms by 5 ^@C5 ^@C, what is the specific heat of the material that the object is made of?