To determine how much heat is released when freezing 47.30 grams of copper at its freezing point of [tex]\( 1,085^{\circ} C \)[/tex], you will use the concept of latent heat of fusion. Latent heat of fusion is the amount of heat energy required to change a substance from the liquid phase to the solid phase (or vice versa) without changing its temperature.
Given:
- Mass of copper, [tex]\( m = 47.30 \)[/tex] grams
- Latent heat of fusion of copper, [tex]\( L = 205.0 \)[/tex] J/g
The formula to calculate the heat released (Q) when a substance changes phase is:
[tex]\[ Q = m \times L \][/tex]
Now substituting the given values:
[tex]\[ Q = 47.30 \text{ grams} \times 205.0 \text{ J/g} \][/tex]
Perform the multiplication:
[tex]\[ Q = 47.30 \times 205.0 = 9696.5 \text{ joules} \][/tex]
So, the amount of heat released to freeze 47.30 grams of copper at its freezing point is [tex]\( 9,696.5 \)[/tex] joules. Since the list of options includes [tex]\( 9,697 \)[/tex] joules (which rounds to our answer), the correct choice is:
[tex]\[ \boxed{9,697 \text{ joules}} \][/tex]
