An object with a mass of $32 g$ is dropped into $560 mL$ of water at $0^@C$ . If the object cools by $80 ^@C$ and the water warms by $6 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-12-05T06:52:29

The specific heat is $=5.49 kJkg^-1K^-1$

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

For the cold water, $Delta T_w=6ºC$

For the object $DeltaT_o=80ºC$

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

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 $m_o=0.032kg$

The mass of the water is $m_w=0.56kg$

$0.032*C_o*80=0.56*4.186*6$

$C_o=(0.56*4.186*6)/(0.032*80)$

$=5.49 kJkg^-1K^-1$

An object with a mass of 32 g32 g is dropped into 560 mL560 mL of water at 0^@C0^@C. If the object cools by 80 ^@C80 ^@C and the water warms by 6 ^@C6 ^@C, what is the specific heat of the material that the object is made of?