Answer:
To calculate the caloric value (in kcal/g) of the Cheetos, you can use the formula:
\[ Q = mc\Delta T \]
Where:
- \( Q \) is the heat absorbed or released,
- \( m \) is the mass of the substance (Cheetos in this case),
- \( c \) is the specific heat of water (assuming the heat is transferred to water), and
- \( \Delta T \) is the change in temperature.
The heat gained by water (\( Q_{\text{water}} \)) is equal to the heat lost by the Cheetos (\( Q_{\text{Cheetos}} \)):
\[ Q_{\text{water}} = Q_{\text{Cheetos}} \]
\[ m_{\text{water}}c_{\text{water}}\Delta T_{\text{water}} = -m_{\text{Cheetos}}c_{\text{Cheetos}}\Delta T_{\text{Cheetos}} \]
Given that \( m_{\text{water}} = 250.0 \, \text{g} \), \( c_{\text{water}} = 4.18 \, \text{J/g}^\circ \text{C} \), \( \Delta T_{\text{water}} = 58 - 21 = 37 \, \text{°C} \), and \( m_{\text{Cheetos}} = 2.8 \, \text{g} \), \( \Delta T_{\text{Cheetos}} = -37 \, \text{°C} \) (negative because heat is lost):
\[ (250.0 \, \text{g})(4.18 \, \text{J/g}^\circ \text{C})(37 \, \text{°C}) = -(2.8 \, \text{g})(c_{\text{Cheetos}})(-37 \, \text{°C}) \]
Now, solve for \( c_{\text{Cheetos}} \), the specific heat of Cheetos.
\[ c_{\text{Cheetos}} = \frac{(250.0 \, \text{g})(4.18 \, \text{J/g}^\circ \text{C})(37 \, \text{°C})}{2.8 \, \text{g} \times 37 \, \text{°C}} \]
\[ c_{\text{Cheetos}} \approx 107 \, \text{J/g}^\circ \text{C} \]
Now, to find the caloric value in kcal/g, convert from joules to kilocalories:
\[ 1 \, \text{cal} = 4.184 \, \text{J} \]
\[ 1 \, \text{kcal} = 1000 \, \text{cal} \]
\[ c_{\text{Cheetos}} \approx \frac{107 \, \text{J/g}^\circ \text{C}}{4.184 \, \text{J/cal}} \times \frac{1 \, \text{cal}}{1000 \, \text{kcal}} \]
\[ c_{\text{Cheetos}} \approx 0.0256 \, \text{kcal/g}^\circ \text{C} \]
Therefore, the caloric value of the Cheetos is approximately \(0.0256 \, \text{kcal/g}^\circ \text{C}\).