To determine how much solid gold is melted when 3,000 Joules of heat is added, we can use the formula that relates the heat absorbed or released, the mass of the substance, and the heat of fusion. The formula is:
\[ Q = m \cdot L_f \]
Where:
- \( Q \) is the heat absorbed or released (in joules),
- \( m \) is the mass of the substance (in kilograms),
- \( L_f \) is the heat of fusion (in joules per kilogram).
Given:
- \( Q = 3000 , \text{J} \)
- \( L_f = 6.30 \times 10^4 , \text{J/kg} \)
We need to find the mass \( m \) of gold that is melted. Rearranging the formula to solve for \( m \):
\[ m = \frac{Q}{L_f} \]
Substituting the known values:
\[ m = \frac{3000 , \text{J}}{6.30 \times 10^4 , \text{J/kg}} \]
Calculating \( m \):
\[ m = \frac{3000}{63000} , \text{kg} \]
\[ m = 0.047619 , \text{kg} \]
Rounding to appropriate significant figures, the mass of the gold that is melted is approximately:
\[ m \approx 0.048 , \text{kg} , \text{or} , 48 , \text{g} \]
Therefore, the amount of gold melted is approximately 48 grams.
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