Question

If it takes 41.72 J to he a piece of gold waiting 18.69 g from 10.0°C to 27°C what is the specific heat of gold

Answer 1

You have to use the equation q=mcΔT and solve for c. (c=q/mΔT)
q=the amount of heat absorbed or released (in this case 41.72)
c=specific heat capacity
m= mass of sample (in this case 18.69g)
ΔT=the change in temperature (in this case 17°C)
when you plug everything in you should bet c=0.1313J/g°C
Therefore, the specific heat capacity of gold is 0.1313J/g°C

I hope this helps. Let me know in the comments if anything is unclear.