An object with a mass of $18 g$ $18 g$ is dropped into $400 m L$ $400 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $150^{\circ} C$ $150 ^@C$ and the water warms by $12^{\circ} C$ $12 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-12-28T00:18:36

$= \frac{7.4311 J}{g \cdot C}$ $=(7.4311J)/(g*C)$

Given the volume of water used as medium to cool down the object, its mass can be derived through the density formula; that is,
$ρ = \frac{m}{V}$ $rho=m/V$
$m = ρ \times V$ $m=rhoxxV$
$m=(1g)/cancel((ml))xx400cancel(ml)$
$m=400gH_2O$
Find the heat absorbed (gained) by the water that increases its temperature by $12^o$ . Knowing the $Cp_w=(4.18J)/(g*C)$ , heat gained can be calculated by:
$Q=mCpDeltaT$
$Q=(400cancel(g))((4.18J)/cancel((g*C)))(12)cancel(^oC)$
$Q=20,064J=20.064kJ$
Since heat lost=heat gained; therefore, the $Cp$ of the material can be computed as;
$Q=mCpDeltaT$
$Cp_(o)=(Q)/(mDeltaT)$
$Cp_(o)=(20,064J)/((18g)(150^oC))$
$Cp_(o)=(7.4311J)/(g*C)$

An object with a mass of 18 g18g18 g is dropped into 400 mL400mL400 mL of water at 0^@C0∘C0^@C. If the object cools by 150 ^@C150∘C150 ^@C and the water warms by 12 ^@C12∘C12 ^@C, what is the specific heat of the material that the object is made of?