An object with a mass of $180 g$ $180 g$ is dropped into $660 m L$ $660 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $36^{\circ} C$ $36 ^@C$ and the water warms by $3^{\circ} C$ $3 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-06-19T05:52:20

The specific heat is $= 1.28 k J k g^{- 1} K^{- 1}$ $=1.28kJkg^-1K^-1$

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

For the cold water, $Delta T_w=3ºC$

For the object $DeltaT_o=36ºC$

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

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

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

Mass of the object is $m_0=0.18kg$

Mass of the water is $m_w=0.66kg$

$0.18*C_o*36=0.66*4.186*3$

$C_o=(0.66*4.186*3)/(0.18*36)$

$=1.28kJkg^-1K^-1$

An object with a mass of 180 g180g180 g is dropped into 660 mL660mL660 mL of water at 0^@C0∘C0^@C. If the object cools by 36 ^@C36∘C36 ^@C and the water warms by 3 ^@C3∘C3 ^@C, what is the specific heat of the material that the object is made of?