To identify the unknown substance based on the experimental calculation of its specific heat capacity, we will use the given data:
- Mass of the substance (m): 12.0 grams
- Heat energy added (Q): 1656 Joules
- Temperature increase (ΔT): 150 degrees Celsius
The formula to calculate the specific heat capacity ([tex]\( c \)[/tex]) is given by:
[tex]\[ Q = mcΔT \][/tex]
We can rearrange this formula to solve for [tex]\( c \)[/tex]:
[tex]\[ c = \frac{Q}{mΔT} \][/tex]
Substituting the given values into this formula, we get:
[tex]\[ c = \frac{1656 \text{ J}}{12.0 \text{ g} \times 150 ^{\circ} \text{C}} \][/tex]
Next, we calculate the denominator:
[tex]\[ 12.0 \text{ g} \times 150 ^{\circ} \text{C} = 1800 \text{ g} \cdot ^{\circ} \text{C} \][/tex]
Thus:
[tex]\[ c = \frac{1656 \text{ J}}{1800 \text{ g} \cdot ^{\circ} \text{C}} \][/tex]
Simplifying this:
[tex]\[ c = 0.92 \text{ J/g} \cdot ^{\circ} \text{C} \][/tex]
Now that we have the calculated specific heat capacity, we can compare it to the given specific heat capacities for various substances:
- Aluminum: [tex]\(0.92 \text{ J/g} \cdot ^{\circ} \text{C} \)[/tex]
- Iron: [tex]\(0.45 \text{ J/g} \cdot ^{\circ} \text{C} \)[/tex]
- Copper: [tex]\(0.39 \text{ J/g} \cdot ^{\circ} \text{C} \)[/tex]
- Glass: [tex]\(0.84 \text{ J/g} \cdot ^{\circ} \text{C} \)[/tex]
The calculated specific heat capacity [tex]\(0.92 \text{ J/g} \cdot ^{\circ} \text{C} \)[/tex] matches precisely with the specific heat capacity of Aluminum.
Therefore, the unknown substance is Aluminum.
