## Conceptual Physics (12th Edition)

a. Find the heat transferred to the water. $$Q = cm \Delta T$$ $$=\frac{1 cal}{g \cdot ^{\circ}C}(50 g)(50^{\circ}C - 22^{\circ}C ) = 1400 cal$$ At 40% efficiency, only 0.40 of the energy in the peanut raises the water temperature. 1400 cal = 0.40 of the total energy in the peanut. Solve for the peanut's energy: $\frac{1400 cal}{0.40}$ = 3500 cal. b. $\frac{3500 cal}{0.6 g} = \frac{5833 cal}{g} = \frac{5.8 kcal}{g} = \frac{5.8 Calories}{g}$.