An object with a mass of $32 g$ is dropped into $480 mL$ of water at $0^@C$ . If the object cools by $80 ^@C$ and the water warms by $4 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-05-18T04:59:29

The specific heat is $=3.14kJkg^-1K^-1$

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

For the cold water, $Delta T_w=4ºC$

For the object $DeltaT_o=80ºC$

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

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

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

$0.032*C_o*80=0.48*4.186*4$

$C_o=(0.48*4.186*4)/(0.032*80)$

$=3.14kJkg^-1K^-1$

An object with a mass of 32 g32 g is dropped into 480 mL480 mL of water at 0^@C0^@C. If the object cools by 80 ^@C80 ^@C and the water warms by 4 ^@C4 ^@C, what is the specific heat of the material that the object is made of?