### Part 1: Calculation of Heat Released
To determine the amount of heat released when a 480 g sample of a substance drops in temperature by [tex]\(7^\circ \mathrm{C}\)[/tex] with a specific heat capacity given by [tex]\( c = \frac{1264 \, \text{J}}{g \cdot {}^\circ \mathrm{C}} \)[/tex]:
1. Determine the given parameters:
- Mass ([tex]\(m\)[/tex]) = 480 g
- Temperature change ([tex]\(\Delta T\)[/tex]) = [tex]\(-7^\circ \mathrm{C}\)[/tex] (since the temperature drops, [tex]\(\Delta T\)[/tex] is negative)
- Specific heat capacity ([tex]\(c\)[/tex]) = 1264 [tex]\(\frac{\text{J}}{g \cdot {}^\circ \mathrm{C}}\)[/tex]
2. Use the heat transfer formula:
[tex]\[
Q = m \cdot c \cdot \Delta T
\][/tex]
3. Substitute the known values into the formula:
[tex]\[
Q = 480 \, \text{g} \cdot 1264 \, \frac{\text{J}}{g \cdot {}^\circ \mathrm{C}} \cdot (-7^\circ \mathrm{C})
\][/tex]
4. Calculate:
[tex]\[
Q = 480 \times 1264 \times (-7) = -4,247,040 \, \text{J}
\][/tex]
Thus, the amount of heat released is [tex]\(-4,247,040\)[/tex] Joules.
### Part 2: Calculation of Specific Heat Capacity
To determine the specific heat capacity of a substance when 2508 g of it increases in temperature by [tex]\(4,051^\circ \mathrm{C}\)[/tex] as it absorbs 3420 Joules of energy:
1. Determine the given parameters:
- Mass ([tex]\(m\)[/tex]) = 2508 g
- Temperature change ([tex]\(\Delta T\)[/tex]) = [tex]\(4,051^\circ \mathrm{C}\)[/tex]
- Energy absorbed ([tex]\(Q\)[/tex]) = 3420 J
2. Use the formula for specific heat capacity:
[tex]\[
c = \frac{Q}{m \cdot \Delta T}
\][/tex]
3. Substitute the known values into the formula:
[tex]\[
c = \frac{3420 \, \text{J}}{2508 \, \text{g} \cdot 4051^\circ \mathrm{C}}
\][/tex]
4. Calculate:
[tex]\[
c = \frac{3420}{2508 \times 4051} \approx 0.000337 \, \frac{\text{J}}{g \cdot {}^\circ \mathrm{C}}
\][/tex]
Thus, the specific heat capacity is approximately [tex]\(0.000337 \frac{\text{J}}{g \cdot {}^\circ \mathrm{C}}\)[/tex].
