An object with a mass of $90 g$ is dropped into $500 mL$ of water at $0^@C$ . If the object cools by $30 ^@C$ and the water warms by $30 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-07-19T16:34:37

The specific heat is $=23.3kJkg^-1K^-1$

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

For the cold water, $Delta T_w=30^@C$

For the object $DeltaT_o=30^@C$

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

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

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

Mass of the object is $m_0=0.090kg$

Mass of the water is $m_w=0.5kg$

$0.09*C_o*30=0.5*4.186*30$

$C_o=(0.5*4.186*30)/(0.09*30)$

$=23.3kJkg^-1K^-1$

An object with a mass of 90 g90 g is dropped into 500 mL500 mL of water at 0^@C0^@C. If the object cools by 30 ^@C30 ^@C and the water warms by 30 ^@C30 ^@C, what is the specific heat of the material that the object is made of?