Calculate the amount of heat (in kJ) required to raise the temperature of an 86.0 g sample of ethanol from 198.0 K to 380.0 K. The specific heat capacity of ethanol is [tex]2.42 \, J/g^{\circ}C[/tex].

A. 60.8 kJ
B. 21.2 kJ
C. 73.6 kJ
D. 28.4 kJ
E. 11.9 kJ



Answer :

To calculate the amount of heat required to raise the temperature of a sample, we can use the formula:

[tex]\[ Q = m \times c \times \Delta T \][/tex]

where:
- [tex]\( Q \)[/tex] is the heat energy (in Joules, J).
- [tex]\( m \)[/tex] is the mass of the sample (in grams, g).
- [tex]\( c \)[/tex] is the specific heat capacity (in J/(g°C)).
- [tex]\( \Delta T \)[/tex] is the change in temperature (in °C or K).

### Step-by-Step Solution

1. Given Data:
- Mass [tex]\( m = 86.0 \, \text{g} \)[/tex]
- Specific heat capacity [tex]\( c = 2.42 \, \text{J/(g°C)} \)[/tex]
- Initial temperature [tex]\( T_\text{initial} = 198.0 \, \text{K} \)[/tex]
- Final temperature [tex]\( T_\text{final} = 380.0 \, \text{K} \)[/tex]

2. Calculate Temperature Change:
Since temperature changes in Kelvin are equivalent to temperature changes in Celsius (°C),
[tex]\[ \Delta T = T_\text{final} - T_\text{initial} = 380.0 \, \text{K} - 198.0 \, \text{K} = 182.0 \, \text{K} \][/tex]

3. Calculate the Heat Energy [tex]\( Q \)[/tex]:
Substitute the values into the formula:
[tex]\[ Q = m \times c \times \Delta T \][/tex]
[tex]\[ Q = 86.0 \, \text{g} \times 2.42 \, \text{J/(g°C)} \times 182.0 \, \text{K} \][/tex]
[tex]\[ Q = 86.0 \times 2.42 \times 182.0 \][/tex]
[tex]\[ Q = 37877.84 \, \text{J} \][/tex]

4. Convert Joules to Kilojoules:
[tex]\[ 1 \, \text{kJ} = 1000 \, \text{J} \][/tex]
Therefore,
[tex]\[ Q = \frac{37877.84 \, \text{J}}{1000} = 37.87784 \, \text{kJ} \][/tex]

### Conclusion
The amount of heat required to raise the temperature of the 86.0 g sample of ethanol from 198.0 K to 380.0 K is approximately [tex]\( 37.88 \, \text{kJ} \)[/tex].

Considering the given options:
- 60.8 kJ
- 21.2 kJ
- 73.6 kJ
- 28.4 kJ
- 11.9 kJ

None of the provided options are correct based on the calculated result of approximately 37.88 kJ.