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 $9^{\circ} C$ $9 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-03-05T14:45:32

The specific heat is $= 3.84 k J k g^{- 1} K^{- 1}$ $=3.84kJkg^-1K^-1$

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

For the cold water, $Delta T_w=9º$

For the metal $DeltaT_o=36º$

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

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

$m_0 C_o*36 = m_w* 4.186 *9$

$0.18*C_o*36=0.66*4.186*9$

$C_o=(0.66*4.186*9)/(0.18*36)$

$=3.84kJkg^-1K^-1$

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 9 ^@C9∘C9 ^@C, what is the specific heat of the material that the object is made of?