An object with a mass of $112 g$ is dropped into $750 mL$ of water at $0^@C$ . If the object cools by $70 ^@C$ and the water warms by $3 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-02-20T10:59:52

The specific heat is $=1.201kJkg^-1K^-1$

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

Let $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=3º$

For the object $DeltaT_o=70º$

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

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

$m_0 C_o*70 = m_w* 4.186 *3$

$0.112*C_o*70=0.75*4.186*3$

$C_o=(0.75*4.186*3)/(0.112*70)$

$=1.201kJkg^-1K^-1$

An object with a mass of 112 g112 g is dropped into 750 mL750 mL of water at 0^@C0^@C. If the object cools by 70 ^@C70 ^@C and the water warms by 3 ^@C3 ^@C, what is the specific heat of the material that the object is made of?