If it takes 41.72 joules to heat a piece of gold weighing 18.69 g from 10.0°C to 27.0°C, what is the specific heat of the gold?

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If it takes 41.72 joules to heat a piece of gold weighing 18.69 g from 10.0°C to 27.0°C, what is the specific heat of the gold?

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Answer 1 · 2016-06-03T17:23:20

$s = 0.131 \frac{J}{g \cdot^{\circ} C}$ $s=0.131J/(g*""^@C)$

Explanation:

The amount of heat ( $q$ $q$ ) absorbed by gold could be expressed as:

$q = m \times s \times Δ T$ $q=mxxsxxDeltaT$

where, $m = 18.69 g$ $m=18.69g$ is the mass of the object,
$s$ $s$ is the specific heat capacity,
and $Δ T = T_{f} - T_{i}$ $DeltaT=T_f-T_i$ is the change on temperature of the sample.

Thus, the specific heat capacity is found by:

$s = \frac{q}{m \times Δ T} = \frac{41.72 J}{18.69 g \times ({27.0}^{\circ} C - {10.0}^{\circ} C)} = 0.131 \frac{J}{g \cdot^{\circ} C}$ $s=q/(mxxDeltaT)=(41.72J)/(18.69gxx(27.0^@C-10.0^@C))=0.131J/(g*""^@C)$

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If it takes 41.72 joules to heat a piece of gold weighing 18.69 g from 10.0°C to 27.0°C, what is the specific heat of the gold?

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