To determine how much heat is released when 25.0 grams of liquid copper at its freezing point of \(1,085^{\circ} C\) changes to solid, we use the concept of latent heat of fusion. The latent heat of fusion (or enthalpy of fusion) is the amount of heat energy required to change 1 gram of a substance from solid to liquid (or liquid to solid) at its melting/freezing point without a change in temperature.
Given:
- Mass of copper, \( m = 25.0 \) grams
- Latent heat of fusion of copper, \( L = 205.0 \, \text{J/g} \)
The heat \( Q \) released during the phase change can be calculated using the formula:
[tex]\[ Q = m \times L \][/tex]
Substituting in the given values:
[tex]\[ Q = 25.0 \, \text{g} \times 205.0 \, \text{J/g} \][/tex]
Let's proceed with the multiplication:
[tex]\[ Q = 5125.0 \, \text{J} \][/tex]
Since the copper is releasing heat as it changes from the liquid to solid phase, the heat released is negative:
[tex]\[ Q = -5125.0 \, \text{J} \][/tex]
Therefore, the amount of heat released is \( -5125.0 \, \text{J} \).
Comparing this result to the provided choices:
- \( -27,130 \, \text{J} \)
- \( -5,130 \, \text{J} \) (This appears to be a minor typographical error and should be \( -5,125.0 \, \text{J} \))
- \( 5,130 \, \text{J} \)
- \( 27,130 \, \text{J} \)
The correct answer is:
[tex]\[ -5,130 \, \text{J} \][/tex]
