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1. What amount of heat is released when the temperature of 480 g of a substance drops by [tex]7^{\circ} C[/tex]?

Given: [tex]c=\frac{1264 \text{ J}}{g^{\circ} C}[/tex]

2. If 2508 g of a substance increases in temperature by [tex]4.051^{\circ} C[/tex] when it absorbs 3.42 kJ of energy, what is the specific heat capacity?


Sagot :

### 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].
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