A 300 kg black bear hibernates in the winter. During hibernation the bear's body temperature drops from 37 to 31°C. How many grams of glucose must the bear metabolize to restore its body temperature to normal? Assume the bear's body is mostly water with a specific heat capacity of 4.2 J/(g . °C) and that all the energy from the combustion of glucose is used to raise the bear's body temperature. (Glucose: $\Delta$H(combustion) = -2,830 kJ/mol, molar mass = 180 g/mol)

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To calculate the energy needed to raise the bear's body temperature, use the formula: $ q = mc\Delta T $. Here, $ m = 300,000 $ g, $ c = 4.2 $ J/(g°C), and $ \Delta T = 6 $°C. This gives $ q = 300,000 \times 4.2 \times 6 = 7,560,000 $ J or 7,560 kJ. The energy from the combustion of glucose is 2,830 kJ/mol. Therefore, the moles of glucose needed is $ 7,560 / 2,830 \approx 2.67 $ mol. Converting moles to grams: $ 2.67 \times 180 \approx 481 $ g.

A 300 Kg Black Bear Hibernates in the Winter Δ\DeltaΔ H(combustion) = -2,830 KJ/mol, Molar Mass = 180 G/mol) A)

A 300 Kg Black Bear Hibernates in the Winter $\Delta$ H(combustion) = -2,830 KJ/mol, Molar Mass = 180 G/mol)
A)