An object with a mass of $160 g$ is dropped into $880 mL$ of water at $0^@C$ . If the object cools by $120 ^@C$ 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 · 2017-02-24T14:34:52

The specific heat is $=2.30kJkg^-1K^-1$

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

For the cold water, $Delta T_w=12º$

For the object $DeltaT_o=120º$

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

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

$m_0 C_o*120 = m_w* 4.186 *12$

$0.160*C_o*120=0.88*4.186*12$

$C_o=(0.88*4.186*12)/(0.160*120)$

$=2.30kJkg^-1K^-1$

An object with a mass of 160 g160 g is dropped into 880 mL880 mL of water at 0^@C0^@C. If the object cools by 120 ^@C120 ^@C and the water warms by 12 ^@C12 ^@C, what is the specific heat of the material that the object is made of?