A 3.0 times 10^2 kg black bear hibernates in the winter. During hibernation the

Question

A 3.0 times 10^2 kg black bear hibernates in the winter. During hibernation the bear's body temperature drops from 37.0 to 31.0 degree C. About 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 4.18 J/(g degree C) and that all the energy from the combustion of glucose is used to raise the bear's body temperature. (Glucose: Delta H_comb = - 2808 kJ/mol, molar mass = 180 g/mol) C_6H_12O_6(g) + 6 O_2(g) rightarrow 6 CO_2(g) + 6 H_2O(t) a. 16 g b. 3.0 kg c. 480 g d. 2.7 g e. 7.5 kg Magnesium metal burns brightly in the air to make a white powder. What are the likely products? a. Mg(H_2O) and MgO_2 b. Mg_3N_2 and MgO c. MgNO d. Mg_2 N_3 and Mg_2O e. MgCO_3 and MgCO Crews of commercial airlines are exposed to a higher level of background cosmic radiation during flight time than they would receive on the ground. A single round-trip, coast-to-coast flight gives a dose of about 0.012 rems. If a flight attendant made 200 such flights in one year, how would his/her annual dose with a normal annual exposure on the ground of 3.6 mSv? (100 rem = 1 Sv) a. a bit less than 10 times more b. a bit more than 100 times more c. a bit more than 10 times more d. bit less than 100 times more e. about the same

Explanation / Answer

(1)

Heat change = m * s * (t2-t1)

q = 3.0 * 102 * 103 * 4.184 * (31.0 - 37.0)

q = - 7.52 * 106 J of energy has been lost.

2808 * 103 J of heat is released by the 1 mol (180 g.) of glucose on combustion

then, 7.52* 106 J of heat is released by 180 * 7.52 * 106 / 2808*103 = 482 g.

(C)

(2)

2 Mg (s) + O2 (g) --------------> 2 MgO (s)

3 Mg (s) + N2 (g) --------------> Mg3N2 (s)

(b)

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