An object with a mass of $120 g$ is dropped into $640 mL$ of water at $0^@C$ . If the object cools by $20 ^@C$ and the water warms by $16 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-07-20T10:30:53

The specific heat is $=17.86kJkg^-1K^-1$

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

For the cold water, $Delta T_w=16^@C$

For the object $DeltaT_o=20^@C$

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

The specific heat of water is $C_w=4.186kJkg^-1K^-1$

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

Mass of the object is $m_0=0.120kg$

Mass of the water is $m_w=0.64kg$

$0.12*C_o*20=0.64*4.186*16$

$C_o=(0.64*4.186*16)/(0.12*20)$

$=17.86kJkg^-1K^-1$

An object with a mass of 120 g120 g is dropped into 640 mL640 mL of water at 0^@C0^@C. If the object cools by 20 ^@C20 ^@C and the water warms by 16 ^@C16 ^@C, what is the specific heat of the material that the object is made of?