An object with a mass of $32 g$ is dropped into $480 mL$ of water at $0^@C$ . If the object cools by $80 ^@C$ and the water warms by $6 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-02-27T08:16:43

The specific heat is $=4.71kJkg^-1K^-1$

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

For the cold water, $Delta T_w=6º$

For the object $DeltaT_o=80º$

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

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

$m_0 C_o*80 = m_w* 4.186 *6$

$0.032*C_o*80=0.48*4.186*6$

$C_o=(0.48*4.186*6)/(0.032*80)$

$=4.71kJkg^-1K^-1$

An object with a mass of 32 g32 g is dropped into 480 mL480 mL of water at 0^@C0^@C. If the object cools by 80 ^@C80 ^@C and the water warms by 6 ^@C6 ^@C, what is the specific heat of the material that the object is made of?