An object with a mass of $21 g$ is dropped into $400 mL$ of water at $0^@C$ . If the object cools by $160 ^@C$ and the water warms by $12 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-11-27T19:25:44

The specific heat of the object is $=5.98 kJkg^-1K^-1$

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

For the cold water, $Delta T_w=12ºC$

For the object $DeltaT_o=160ºC$

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

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

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

The mass of the object is $m_o=0.021kg$

The mass of the water is $m_w=0.4kg$

$0.021*C_o*160=0.40*4.186*12$

$C_o=(0.40*4.186*12)/(0.021*160)$

$=5.98 kJkg^-1K^-1$

An object with a mass of 21 g21 g is dropped into 400 mL400 mL of water at 0^@C0^@C. If the object cools by 160 ^@C160 ^@C and the water warms by 12 ^@C12 ^@C, what is the specific heat of the material that the object is made of?