A \(\pu{1.0 g}\) sample of a cashew was burned in a calorimeter containing \(\pu{1000}\). \(\pu{g}\) of water, and the temperature of the water changed from \(20.0^\circ \pu{C}\) to \(25.0^\circ \pu{C}\). In another experiment, a 3.0 g sample of a marshmellow was burned in a calorimeter containing \(2000\). \(\pu{g}\) of water, and the temperature of the water changed from \(25.0^\circ \pu{C}\) to \(30.0^\circ \pu{C}\). Based on the data, which of the following can be concluded about the energy content for \(\pu{1.0 g}\) of each of the two substances? (The specific heat of water is \(\pu{4.2 J}/(\pu{g\cdot ^\circ}\pu{C})\).)

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We can calculate the amount of heat energy transferred by each food item:

$$ q_{\text{cashew}}= q_{\text{water}} = mc\Delta T $$ $$ = (1000)(4.2)(25-20) $$ $$ \approx \pu{20,000 J} $$ $$ q_{\text{marshmellow}}= q_{\text{water}} = mc\Delta T $$ $$ = (2000)(4.2)(30-25) $$ $$ \approx \pu{40,000 J} $$

Note that the \(\pu{20,000 J}\) was transferred from \(\pu{1.0}\) grams of cashew, whereas the \(\pu{40,000 J}\) was transferred from \(\pu{3.0}\) grams of marshmellow.

The marshmellow transfers \(\dfrac{40{,}000}{3} \pu{J}\) of energy per gram, which is significantly less than \(\pu{20,000 J}\).