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How many kilojoules of energy are required to raise the temperature of 49.8 g of water from 22.15°C to 76.90°C?

Sagot :

Sure, let's break this down step-by-step to find how many kilojoules of energy are required:

1. Identify the given values:
- Mass of the water ([tex]\(m\)[/tex]): 49.8 grams
- Specific heat capacity of water ([tex]\(c\)[/tex]): 4.184 J/g°C
- Initial temperature ([tex]\(T_i\)[/tex]): 22.15°C
- Final temperature ([tex]\(T_f\)[/tex]): 76.90°C

2. Calculate the change in temperature ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T = T_f - T_i = 76.90°C - 22.15°C = 54.75°C \][/tex]

3. Use the formula for heat energy (q):
[tex]\[ q = m \times c \times \Delta T \][/tex]
Substituting the known values:
[tex]\[ q = 49.8 \, \text{g} \times 4.184 \, \text{J/g°C} \times 54.75°C \][/tex]

4. Calculate the energy required in joules:
[tex]\[ q \approx 49.8 \times 4.184 \times 54.75 = 11407.8852 \, \text{J} \][/tex]

5. Convert the energy from joules to kilojoules:
[tex]\[ 1 \, \text{kJ} = 1000 \, \text{J} \][/tex]
[tex]\[ q = \frac{11407.8852 \, \text{J}}{1000} \approx 11.4078852 \, \text{kJ} \][/tex]

Therefore, the energy required to raise the temperature of 49.8 grams of water from 22.15°C to 76.90°C is approximately 11.4 kJ.