Answer :
To solve this problem, we need to calculate the amount of heat energy required to raise the temperature of water. We'll use the formula for heat energy, which is given by:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat energy,
- [tex]\( m \)[/tex] is the mass of the substance,
- [tex]\( c \)[/tex] is the specific heat capacity,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
Step-by-Step Solution:
1. Mass of water ([tex]\( m \)[/tex]):
The mass of water given is 28 grams (g).
2. Specific heat capacity of water ([tex]\( c \)[/tex]):
The specific heat capacity of water is 4.18 Joules per gram per degree Celsius (J/g°C).
3. Initial and final temperatures ([tex]\( T_i \)[/tex] and [tex]\( T_f \)[/tex]):
The initial temperature of the water is 250 Kelvin (K) and the final temperature is 279 Kelvin (K).
4. Convert temperatures from Kelvin to Celsius:
We need to convert the temperatures from Kelvin to Celsius because the specific heat capacity is given in J/g°C.
The conversion formula is:
[tex]\[ T_C = T_K - 273.15 \][/tex]
- Initial temperature in Celsius ([tex]\( T_i \)[/tex]):
[tex]\[ T_i = 250 \, K - 273.15 = -23.15 \, °C \][/tex]
- Final temperature in Celsius ([tex]\( T_f \)[/tex]):
[tex]\[ T_f = 279 \, K - 273.15 = 5.85 \, °C \][/tex]
5. Calculate the temperature change ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_f - T_i = 5.85 \, °C - (-23.15 \, °C) = 5.85 \, °C + 23.15 \, °C = 29 \, °C \][/tex]
6. Calculate the heat energy ([tex]\( Q \)[/tex]):
Now apply the values into the heat energy formula:
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = 28 \, g \times 4.18 \, \dfrac{J}{g°C} \times 29 \, °C \][/tex]
[tex]\[ Q = 28 \times 4.18 \times 29 \][/tex]
[tex]\[ Q = 3396.56 \, J \][/tex]
Therefore, the heat energy required is approximately 3,397 Joules.
So, the correct answer is:
c. 3,397 Joules
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat energy,
- [tex]\( m \)[/tex] is the mass of the substance,
- [tex]\( c \)[/tex] is the specific heat capacity,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
Step-by-Step Solution:
1. Mass of water ([tex]\( m \)[/tex]):
The mass of water given is 28 grams (g).
2. Specific heat capacity of water ([tex]\( c \)[/tex]):
The specific heat capacity of water is 4.18 Joules per gram per degree Celsius (J/g°C).
3. Initial and final temperatures ([tex]\( T_i \)[/tex] and [tex]\( T_f \)[/tex]):
The initial temperature of the water is 250 Kelvin (K) and the final temperature is 279 Kelvin (K).
4. Convert temperatures from Kelvin to Celsius:
We need to convert the temperatures from Kelvin to Celsius because the specific heat capacity is given in J/g°C.
The conversion formula is:
[tex]\[ T_C = T_K - 273.15 \][/tex]
- Initial temperature in Celsius ([tex]\( T_i \)[/tex]):
[tex]\[ T_i = 250 \, K - 273.15 = -23.15 \, °C \][/tex]
- Final temperature in Celsius ([tex]\( T_f \)[/tex]):
[tex]\[ T_f = 279 \, K - 273.15 = 5.85 \, °C \][/tex]
5. Calculate the temperature change ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_f - T_i = 5.85 \, °C - (-23.15 \, °C) = 5.85 \, °C + 23.15 \, °C = 29 \, °C \][/tex]
6. Calculate the heat energy ([tex]\( Q \)[/tex]):
Now apply the values into the heat energy formula:
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = 28 \, g \times 4.18 \, \dfrac{J}{g°C} \times 29 \, °C \][/tex]
[tex]\[ Q = 28 \times 4.18 \times 29 \][/tex]
[tex]\[ Q = 3396.56 \, J \][/tex]
Therefore, the heat energy required is approximately 3,397 Joules.
So, the correct answer is:
c. 3,397 Joules
