An object with a mass of $64 g$ is dropped into $320 mL$ of water at $0^@C$ . If the object cools by $16 ^@C$ and the water warms by $6 ^@C$ , what is the specific heat of the material that the object is made of?

Question

1451 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-24T13:59:56

The specific heat is $=7.85kJkg^-1k^-1$

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

For the cold water, $Delta T_w=6ºC$

For the object $DeltaT_o=16ºC$

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

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

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

$0.064*C_o*16=0.32*4.186*6$

$C_o=(0.32*4.186*6)/(0.064*16)$

$=7.85kJkg^-1k^-1$

An object with a mass of 64 g64 g is dropped into 320 mL320 mL of water at 0^@C0^@C. If the object cools by 16 ^@C16 ^@C and the water warms by 6 ^@C6 ^@C, what is the specific heat of the material that the object is made of?