Get expert insights and community-driven knowledge on IDNLearn.com. Our platform is designed to provide accurate and comprehensive answers to any questions you may have.

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.