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QuestionCalculate the energy required to heat 368.0 g of aluminum from 0.3∘C0.3^{\circ} \mathrm{C} to 18.5∘C18.5^{\circ} \mathrm{C}. Assume the specific heat capacity of aluminum under these conditions is 0.903 J⋅ g−1⋅ K−10.903 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}. Be sure your answer has the correct number of significant digits.

Studdy Solution

STEP 1

1. The mass of aluminum is 368.0 g 368.0 \, \text{g} .
2. The initial temperature is 0.3∘C 0.3^{\circ} \mathrm{C} .
3. The final temperature is 18.5∘C 18.5^{\circ} \mathrm{C} .
4. The specific heat capacity of aluminum is 0.903 J⋅g−1⋅K−1 0.903 \, \mathrm{J} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1} .
5. The formula to calculate energy is q=m⋅c⋅ΔT q = m \cdot c \cdot \Delta T .

STEP 2

1. Calculate the temperature change (ΔT\Delta T).
2. Use the formula for heat energy to calculate the energy required.
3. Ensure the answer has the correct number of significant figures.

STEP 3

Calculate the temperature change (ΔT\Delta T).
ΔT=Tfinal−Tinitial=18.5∘C−0.3∘C=18.2∘C\Delta T = T_{\text{final}} - T_{\text{initial}} = 18.5^{\circ} \mathrm{C} - 0.3^{\circ} \mathrm{C} = 18.2^{\circ} \mathrm{C}

STEP 4

Use the formula for heat energy:
q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta T
Substitute the known values:
q=368.0 g×0.903 J⋅g−1⋅K−1×18.2 Kq = 368.0 \, \text{g} \times 0.903 \, \mathrm{J} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1} \times 18.2 \, \mathrm{K}

STEP 5

Calculate the energy required:
q=368.0×0.903×18.2=6044.8416 Jq = 368.0 \times 0.903 \times 18.2 = 6044.8416 \, \mathrm{J}

STEP 6

Round the answer to the correct number of significant figures. The least number of significant figures in the given data is 4 (from 368.0 g 368.0 \, \text{g} ).
q=6044.8416 J≈6044 Jq = 6044.8416 \, \mathrm{J} \approx 6044 \, \mathrm{J}
The energy required to heat the aluminum is:
6044 J\boxed{6044 \, \mathrm{J}}

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