🤓 Based on our data, we think this question is relevant for Professor Dixon's class at UCF.

Assuming water is at room temperature (25°C) at the beginning. Specific heat of water is 4.18 J/g°C

$\overline{)\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{mC}\mathbf{\u2206}\mathbf{T}}\phantom{\rule{0ex}{0ex}}\mathbf{(}\mathbf{14}\mathbf{.}\mathbf{5}\mathbf{}\overline{)\mathbf{kJ}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{J}}{\mathbf{1}\mathbf{}\overline{)\mathbf{kJ}}}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{485}\mathbf{}\overline{)\mathbf{g}}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{18}\mathbf{}\frac{\mathbf{J}}{\overline{)\mathbf{g}}\mathbf{\xb0}\mathbf{C}}\mathbf{)}\mathbf{(}{\mathbf{T}}_{\mathbf{f}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{25}\mathbf{\xb0}\mathbf{C}\mathbf{)}\phantom{\rule{0ex}{0ex}}{\mathbf{T}}_{\mathbf{f}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{25}\mathbf{\xb0}\mathbf{C}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{14}\mathbf{.}\mathbf{5}\mathbf{}\overline{)\mathbf{kJ}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{J}}{\mathbf{1}\mathbf{}\overline{)\mathbf{kJ}}}\mathbf{\right)}\mathbf{}}{\mathbf{(}\mathbf{485}\mathbf{}\overline{)\mathbf{g}}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{18}\mathbf{}\frac{\overline{)\mathbf{J}}}{\overline{)\mathbf{g}}\mathbf{\xb0}\mathbf{C}}\mathbf{)}}\phantom{\rule{0ex}{0ex}}{\mathbf{T}}_{\mathbf{f}}\mathbf{}\mathbf{-}\mathbf{25}\mathbf{\xb0}\mathbf{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{\xb0}\mathbf{C}$

If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?

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