If [tex][tex]$1.9 \text{ kJ}$[/tex][/tex] of heat is transferred to [tex][tex]$96 \text{ g}$[/tex][/tex] aluminum at [tex][tex]$113^{\circ} \text{C}$[/tex][/tex], what would the new temperature of the aluminum be? (The specific heat capacity of aluminum is [tex][tex]$0.897 \text{ J/g} \cdot{ }^{\circ} \text{C}$[/tex][/tex].)



Answer :

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.