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?

Answer 1

#s=0.131J/(g*""^@C)#

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

#q=mxxsxxDeltaT#

where, #m=18.69g# is the mass of the object,
#s# is the specific heat capacity,
and #DeltaT=T_f-T_i# is the change on temperature of the sample.

Thus, the specific heat capacity is found by:

#s=q/(mxxDeltaT)=(41.72J)/(18.69gxx(27.0^@C-10.0^@C))=0.131J/(g*""^@C)#

Thermochemistry | Enthalpy and Calorimetry.

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Answer 2

Kai Bowman

The specific heat of gold is 0.129 J/g°C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.