To determine the mass of water that can be heated from 6.8°C to 32.8°C by adding 3,884.2 joules of energy, we can use the formula for heat energy:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where [tex]\( Q \)[/tex] is the heat energy (in joules), [tex]\( m \)[/tex] is the mass (in grams), [tex]\( c \)[/tex] is the specific heat capacity of water, and [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
Given:
- [tex]\( Q = 3,884.2 \, \text{J} \)[/tex]
- [tex]\( c = 4.186 \, \text{J/g°C} \)[/tex] (specific heat capacity of water)
- Initial temperature [tex]\( T_1 = 6.8 \, °C \)[/tex]
- Final temperature [tex]\( T_2 = 32.8 \, °C \)[/tex]
First, calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_2 - T_1 = 32.8 \, °C - 6.8 \, °C = 26.0 \, °C \][/tex]
Next, rearrange the formula to solve for the mass [tex]\( m \)[/tex]:
[tex]\[ m = \frac{Q}{c \cdot \Delta T} \][/tex]
Now, substitute the known values into the equation:
[tex]\[ m = \frac{3,884.2 \, \text{J}}{4.186 \, \text{J/g°C} \times 26.0 \, °C} \][/tex]
[tex]\[ m = \frac{3,884.2}{4.186 \times 26.0} \][/tex]
[tex]\[ m = \frac{3,884.2}{108.836} \][/tex]
[tex]\[ m \approx 35.69 \, \text{g} \][/tex]
Therefore, the mass of water that can be heated from 6.8°C to 32.8°C by the addition of 3,884.2 joules of energy is approximately 35.69 grams.
Answer:
[tex]\[ 35.69 \][/tex]
units: [tex]\[ \text{g} \][/tex]
