If aluminum absorbs 1250 J of energy, increasing the temperature from [tex]\( 25^{\circ}C \)[/tex] to [tex]\( 35^{\circ}C \)[/tex], what is the mass of the aluminum? (Specific heat of [tex]\( Al = 0.903 \, J/g^{\circ}C \)[/tex])



Answer :

Certainly! Here is a detailed, step-by-step solution to solve for the mass of aluminum under the given conditions:

1. Identify the given values:
- Energy absorbed by aluminum, [tex]\( Q \)[/tex]: 1250 Joules (J)
- Initial temperature, [tex]\( T_{\text{initial}} \)[/tex]: [tex]\( 25^{\circ}C \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} \)[/tex]: [tex]\( 35^{\circ}C \)[/tex]
- Specific heat capacity of aluminum, [tex]\( c \)[/tex]: [tex]\( 0.903 \, \text{J/g}^{\circ}\text{C} \)[/tex]

2. Calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
- [tex]\( \Delta T = T_{\text{final}} - T_{\text{initial}} \)[/tex]
- [tex]\( \Delta T = 35^{\circ}C - 25^{\circ}C = 10^{\circ}C \)[/tex]

3. Use the formula for heat absorption:
- The formula for heat absorption is [tex]\( Q = mc\Delta T \)[/tex]
- Here, [tex]\( Q \)[/tex] is the energy absorbed, [tex]\( m \)[/tex] is the mass of the substance, [tex]\( c \)[/tex] is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.

4. Rearrange the formula to solve for mass ([tex]\( m \)[/tex]):
- To solve for mass, rearrange the formula to [tex]\( m = \frac{Q}{c\Delta T} \)[/tex]

5. Substitute the known values into the rearranged formula:
- [tex]\( m = \frac{1250 \, \text{J}}{0.903 \, \text{J/g}^{\circ}\text{C} \times 10^{\circ}C} \)[/tex]

6. Perform the calculation:
- [tex]\( m = \frac{1250}{0.903 \times 10} \)[/tex]
- [tex]\( m = \frac{1250}{9.03} \)[/tex]
- [tex]\( m \approx 138.43 \, \text{g} \)[/tex]

So, after absorbing 1250 Joules of energy, and given the specific heat capacity of aluminum is [tex]\( 0.903 \, \text{J/g}^{\circ}\text{C} \)[/tex], the mass of the aluminum is approximately [tex]\( 138.43 \)[/tex] grams.

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