Answer:
The heat energy ([tex]\( q \)[/tex]) required to raise the temperature of 10.9 grams of water from -98.3 °C to -26.8 °C is 3.261 kJ.
Explanation:
Firstly, it's essential to calculate the change in temperature, denoted as [tex]\( \Delta T \)[/tex].
[tex]$\begin{align*}\Delta T &= \text{Final temperature} - \text{Initial temperature} \\ &= -26.8 \, ^\circ C - (-98.3 \, ^\circ C) \\ &= 71.5 \, ^\circ C\end{align*}[/tex]
Next, we calculate the heat energy using the specific heat capacity of water. For water, the specific heat capacity ([tex]\( c \)[/tex]) is 4.184 J/g·°C, which can also be expressed as 0.004184 kJ/g·°C.
[tex]$\begin{align*}q &= m \times c \times \Delta T \\ &= 10.9 \, \text{g} \times 0.004184 \, \text{kJ/g}\cdot ^\circ\text{C} \times 71.5 \, ^\circ\text{C} \\ \therefore q &= \boxed{3.261 \, \text{kJ}}\end{align*}[/tex]
