An object with a mass of $64 g$ $64 g$ is dropped into $320 m L$ $320 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $16^{\circ} C$ $16 ^@C$ and the water warms by $4^{\circ} C$ $4 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-03-08T11:02:54

The specific heat is $= 5.23 k J k g^{- 1} K^{- 1}$ $=5.23kJkg^-1K^-1$

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water, $Delta T_w=4º$

For the object $DeltaT_o=16º$

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

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

$m_0 C_o*16 = m_w* 4.186 *4$

$0.064*C_o*16=0.32*4.186*4$

$C_o=(0.32*4.186*4)/(0.064*16)$

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

An object with a mass of 64 g64g64 g is dropped into 320 mL320mL320 mL of water at 0^@C0∘C0^@C. If the object cools by 16 ^@C16∘C16 ^@C and the water warms by 4 ^@C4∘C4 ^@C, what is the specific heat of the material that the object is made of?