An object with a mass of $180 g$ is dropped into $660 mL$ of water at $0^@C$ . If the object cools by $72 ^@C$ and the water warms by $3 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-07-18T14:07:48

The specific heat is $=0.64kJkg^-1K^-1$

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

For the cold water, $Delta T_w=3^@C$

For the object $DeltaT_o=72^@C$

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

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

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

Mass of the object is $m_0=0.180kg$

Mass of the water is $m_w=0.66kg$

$0.18*C_o*72=0.66*4.186*3$

$C_o=(0.66*4.186*3)/(0.18*72)$

$=0.64kJkg^-1K^-1$

An object with a mass of 180 g180 g is dropped into 660 mL660 mL of water at 0^@C0^@C. If the object cools by 72 ^@C72 ^@C and the water warms by 3 ^@C3 ^@C, what is the specific heat of the material that the object is made of?