An object with a mass of $20 g$ is dropped into $150 mL$ of water at $0^@C$ . If the object cools by $90 ^@C$ and the water warms by $15 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-05-30T05:24:39

The specific heat is $=5.23kJkg^-1K^-1$

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

For the cold water, $Delta T_w=15ºC$

For the object $DeltaT_o=90ºC$

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

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

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

$0.020*C_o*90=0.15*4.186*15$

$C_o=(0.15*4.186*15)/(0.020*90)$

$=5.23kJkg^-1K^-1$

An object with a mass of 20 g20 g is dropped into 150 mL150 mL of water at 0^@C0^@C. If the object cools by 90 ^@C90 ^@C and the water warms by 15 ^@C15 ^@C, what is the specific heat of the material that the object is made of?