Q:

Find out of the friend is getting a good deal

I have a strange chemistry question that I've never came across before, and I have no idea about how to solve it. 

Your friend comes across a good deal to purchase a gold ring.  She asks you for advice for you to test the ring.  The ring has a mass of 4.54 g. When the ring is heated with 94.8 J of energy, temperature rises from 23.0 °C to 47.5 °C.  The gold specific heat capacity is 0.1291 J/g. Is she getting a good deal? 

So I did the equation to solve for Q which is 14.35 J, and I have all the numbers the (considering they are given) but I have no idea how to actually calculate how much of this ring is pure gold.  
I came here as a last resort. 

A:

The jewelry gold usually comes historically rated in grades 14 or 18 karats, where 24 means pure gold.  The rest is mostly silver. 
Find out specific heat capacity for the complementary metals like silver or copper, so you could roughly estimate via interpolation. 
Another method would be pyknometry, determining the ring density by differential weighting the mass of displaced water - old good method of Archimedes.  Gold has almost double density, compared to other expected metals. 
Edit:  list of densities:

Gold 19.3.  
Silver 10.5. 

If linearly interpolated by mass fraction:.     

14 karat gold  15.63. 
18 karat gold  17.1

A:

I agree with Robert Ochinski (see above comment) that this question, as posed, is not answerable, because the price of the ring was not given.  Yet, we can assume what amount of gold in the ring by given data.  The specific heat capacity of ring ($c_{p(\text{ring})}$) can be calculated using the equation, $\Delta Q = mc_p\Delta T$. Thus, substituting given values:
$$c_{p(\text{ring})}= \frac{\pu{94.8 J}}{\pu{4.54 g} \times \left(47.5-23.0 \right)\pu{^{\circ}C}}= \pu{0.852 Jg^{-1}{^{\circ}C^{-1}}}$$ 
This indicate that the ring is not in pure gold, and the specific heat capacity of the mixing metal should be greater than $\pu{0.852 Jg^{-1}{^{\circ}C^{-1}}}$. This makes only mixable metal be aluminum with $c_{p(\ce{Al})}$ of $\pu{0.91 Jg^{-1}{^{\circ}C^{-1}}}$.  Other disregarded metals and metalloids are $\ce{Li}$ ($c_{p(\ce{Li})}=\pu{3.57 Jg^{-1}{^{\circ}C^{-1}}}$), $\ce{Be}$ ($c_{p(\ce{Be})}=\pu{1.83 Jg^{-1}{^{\circ}C^{-1}}}$), $\ce{Na}$ ($c_{p(\ce{Na})}=\pu{1.21 Jg^{-1}{^{\circ}C^{-1}}}$), and $\ce{Mg}$ ($c_{p(\ce{Mg})}=\pu{1.05 Jg^{-1}{^{\circ}C^{-1}}}$), all of which are unstable under ambient conditions (Data are from: Engineering Toolbox).  Keep in mind that well know metals used to make alloys with gold such as $\ce{Ag}$ and $\ce{Cu}$ have specific heat capacities less than $\pu{0.50 Jg^{-1}{^{\circ}C^{-1}}}$. Accordingly, we can set the two equations:
$$m_\ce{Au} + m_\ce{Al} = 4.54  \enspace \enspace \enspace \enspace (1)$$
$$ 0.129 \left(\frac{m_\ce{Au}}{m_\ce{Au}+m_\ce{Al}}\right) + 0.91 \left(\frac{m_\ce{Al}}{m_\ce{Au}+m_\ce{Al}}\right)   = 0.852 \enspace \enspace \enspace \enspace (2)$$
$$ 0.129 \left(\frac{m_\ce{Au}}{4.54}\right) + 0.91 \left(\frac{m_\ce{Al}}{4.54}\right)   = 0.852 \enspace \enspace \enspace \enspace \enspace (3)$$
Where $m_\ce{Au}$ is mass of $\ce{Au}$ in the ring and $m_\ce{Al}$ is mass of $\ce{Al}$ contributed to the ring (all in $\pu{g}$).  The equation (2) is derived from assuming heat did not loose when heating the mixture and each metal increase the temperature relevant to their specific heat capacities (Thermtest). 
When eq (3) is simplified, it gives:
$$ 129 m_\ce{Au} + 910 m_\ce{Al} = 3868 \enspace \enspace \enspace \enspace \enspace (4)$$
Using eq (1) and (4), solve for $m_\ce{Au}$:
$$ (910-129) m_\ce{Au}  = 4.54\times910-3868 $$
$$ \therefore m_\ce{Au}  = \frac{4.54\times910-3868}{910-129} = 0.337$$
Thus, the ring contains only $\pu{0.337 g}$ of gold.  The rest is believed to be aluminum.  Today's gold price is USD 42.13/g (Gold Price).  Accordingly, the ring contains USD 14.20 worth of gold.  I'd say if your friend spent more than $40/- for the ring today (giving some space for labor), she's not having a good deal! :-)