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

A. [tex]$135^{\circ} \text{C}$[/tex]
B. [tex]$22^{\circ} \text{C}$[/tex]
C. [tex]$114^{\circ} \text{C}$[/tex]
D. [tex]$113^{\circ} \text{C}$[/tex]


Sagot :

To find the new temperature of the aluminum after transferring 1.9 kJ of heat, we can follow these steps:

1. Convert the heat transfer from kJ to J:
- The given heat transfer is 1.9 kJ.
- Since 1 kJ = 1000 J, we multiply:
[tex]\[ 1.9 \, \text{kJ} \times 1000 \, \left(\frac{\text{J}}{\text{kJ}}\right) = 1900 \, \text{J} \][/tex]

2. Write down the formula for heat transfer:
- The formula for heat transfer is given by:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat transferred (in Joules),
- [tex]\( m \)[/tex] is the mass (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity (in J/g°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).

3. Rearrange the formula to solve for the change in temperature, [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{Q}{m \cdot c} \][/tex]

4. Substitute the known values into the formula:
- [tex]\( Q = 1900 \, \text{J} \)[/tex]
- [tex]\( m = 96 \, \text{g} \)[/tex]
- [tex]\( c = 0.897 \, \text{J/g°C} \)[/tex]
[tex]\[ \Delta T = \frac{1900 \, \text{J}}{96 \, \text{g} \times 0.897 \, \text{J/g°C}} \][/tex]
[tex]\[ \Delta T \approx \frac{1900}{86.112} \][/tex]
[tex]\[ \Delta T \approx 22.06^{\circ}C \][/tex]

5. Determine the new temperature:
- The initial temperature of the aluminum is [tex]\( 113^{\circ}C \)[/tex].
- Add the change in temperature to the initial temperature to find the new temperature:
[tex]\[ \text{New temperature} = 113^{\circ}C + 22.06^{\circ}C \approx 135.06^{\circ}C \][/tex]

Considering the given options, the closest answer is:
A. [tex]\( 135^{\circ}C \)[/tex]