Certainly! Let's solve this step-by-step:
1. Given Values:
- Amount of heat transferred ([tex]\( q \)[/tex]): [tex]\( 1.9 \)[/tex] kJ
- Mass of aluminum ([tex]\( m \)[/tex]): [tex]\( 96 \)[/tex] g
- Initial temperature of aluminum ([tex]\( T_{\text{initial}} \)[/tex]): [tex]\( 113 \)[/tex]°C
- Specific heat capacity of aluminum ([tex]\( c \)[/tex]): [tex]\( 0.897 \)[/tex] J/(g·°C)
2. Convert heat transferred from kilojoules to joules:
[tex]\[
q = 1.9 \, \text{kJ} \times 1000 \, \left(\frac{\text{J}}{\text{kJ}}\right) = 1900 \, \text{J}
\][/tex]
3. Use the formula to calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[
q = m \times c \times \Delta T
\][/tex]
We need to solve for [tex]\( \Delta T \)[/tex]:
[tex]\[
\Delta T = \frac{q}{m \times c}
\][/tex]
4. Substitute the known values into the formula:
[tex]\[
\Delta T = \frac{1900 \, \text{J}}{96 \, \text{g} \times 0.897 \, \frac{\text{J}}{\text{g} \cdot \, °\text{C}}}
\][/tex]
[tex]\[
\Delta T \approx 22.064 \, °\text{C}
\][/tex]
5. Calculate the new temperature of the aluminum:
[tex]\[
T_{\text{new}} = T_{\text{initial}} + \Delta T
\][/tex]
[tex]\[
T_{\text{new}} = 113 \,°\text{C} + 22.064 \,°\text{C}
\][/tex]
[tex]\[
T_{\text{new}} \approx 135.064 \, °\text{C}
\][/tex]
So, the new temperature of the aluminum after [tex]\( 1.9 \)[/tex] kJ of heat is transferred to it would be approximately [tex]\( 135.064 \)[/tex]°C.
