An object with a mass of $15 g$ is dropped into $250 mL$ of water at $0^@C$ . If the object cools by $120 ^@C$ and the water warms by $9 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2018-04-30T20:22:27

From your data we can assume the object lost heat and the water gained heat in equilibrating.

Moreover, I am assuming:

$rho_(H_2O) = (1.0"g")/"mL"$

Recall,

$q = mC_sDeltaT$

In this case, the heat gained by the water is supplied by the body. Accordingly, we can rearrange the equation, such that,

$-m_xC_xDeltaT = m_(H_2O)C_(H_2O)DeltaT_(H_2O)$

Hence,

$=> C_x = (-m_(H_2O)C_(H_2O)DeltaT_(H_2O))/(mxDeltaT_x) approx (5.23"J")/("g"*°"C")$

is the specific heat of the body of mass we dropped into the water.

An object with a mass of 15 g15 g is dropped into 250 mL250 mL of water at 0^@C0^@C. If the object cools by 120 ^@C120 ^@C and the water warms by 9 ^@C9 ^@C, what is the specific heat of the material that the object is made of?