Answer :
To calculate the energy required to heat the water, we can use the formula for heat transfer:
[tex]q = m \times c \times \Delta T[/tex]
Where:
- [tex]q[/tex] is the heat energy (in joules, J) that is required.
- [tex]m[/tex] is the mass of the water (in grams, g). Since the question provides the mass in milligrams (mg), we need to convert it to grams: [tex]382.0 \ mg = 0.382 \ g[/tex].
- [tex]c[/tex] is the specific heat capacity of water, which is given as [tex]4.18 \ \text{J}\cdot\text{g}^{-1}\cdot\text{K}^{-1}[/tex].
- [tex]\Delta T[/tex] is the change in temperature (in Celsius, °C or Kelvin, K). Celsius and Kelvin can be interchangeable for temperature changes, so:
[tex]\Delta T = 38.8 \, ^\circ\text{C} - 28.7 \, ^\circ\text{C} = 10.1 \, ^\circ\text{C}[/tex]
Now, plug in these values:
[tex]q = 0.382 \, \text{g} \times 4.18 \, \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1} \times 10.1 \, ^\circ\text{C}[/tex]
[tex]q = 16.134636 \, \text{J}[/tex]
Rounding the answer to 3 significant digits, we get:
[tex]q \approx 16.1 \, \text{J}[/tex]
Therefore, the energy required to heat 382.0 mg of water from 28.7°C to 38.8°C is approximately 16.1 Joules.
