I. A, B, C, and E are true statements, whereas D is false.
II. Answers A, B and E are true statements, whereas C and D are false.
1. The extra heat energy stored in his body at the end of the 30 min period is: [3 oC× 70 kg× 3.47 kJ/(oC kg)] = 729 kJ.
2. A minimal metabolic rate is assumed to be just about BMR: 300 kJ/hour or 5 kJ/min or 83 Watts. If the heat loss is assumed to be zero, because all the heat produced in 30 min is stored, it explains an accumulation of no more than: (5× 30 min) = 150 kJ. This does not explain the rise in heat energy stored (729 kJ).
3. Only dramatic shivering during the abrupt rise in temperature can explain the large heat energy store.
4. The heat capacity of the body is 3.47 kJ/kg and oC, and that of water is 1 kcal or 4.2 kJ/kg and oC. Intake of 1 L of ice water corresponds to a maximum of (40 oC × 4.2) kJ = 168 kJ. The body weight multiplied with the heat capacity of the body show that (70 kg × 3.47) = 243 kJ must be eliminated from the body in order to reduce the body temperature of the patient one oC. Accordingly, the ice water can reduce the temperature of the patient by 168/243 = 0.7 oC.
5. Evaporation of sweat water (V°sweat in g/min) implies a loss of energy (QE J/min) according to the equation: QE (J/min) = 2436 (J/g)× V°sweat (g/min) = (2436 × 8 ml/min)10-3 = 19.5 kJ/min. With a water evaporation of (32 × 0.25=) 8 ml/min and 19.5 kJ per min, it will take (729/19.5) = 37 minutes in order to eliminate the accumulated heat energy. - The calculation illustrates the fact that evaporation of only 25% of a sweat volume of about 1 l (37 min and 32 ml/min = 1187 ml) eliminates much more heat energy than ingestion of one l of ice water.
1. The metabolic rate is [400* 17.5 + 100* 39 + 100* 17] = 12 600 kJ = 12.6 MJ/day.
2. The metabolic water formed by oxidation of the food is: 32 g/MJ multiplied by 12,600 kJ. This water volume is 403 g or ml. This amount explains the discrepancy between 2100 and 1700 ml of water.
The old man was in a state of severe hypothermia. The gradual rewarming of his body in the chapel has been an adequate treatment.
The core temperature of the victim at the admission to hospital was below 32oC, since the hypothermic victim was in coma, and probably above 30o C, since he recovered spontaneously, and avoided death from ventricular fibrillation.
1. Cshell = MR/(Tcore - Tshell). Cshell = 80/(37 - 33) = 20 W / oC.
2. Cshell = MR/( Tcore - Tshell ). Cshell = (80 * 7)/40 - 34) = 93 W / oC.
3. Preoptic heat sensors in the hypothalamus show dynamic gain, so with increasing tissue temperature, the neural activity of the sensors increases linearly. These heat sensors stimulate effectors for heat loss, and a cutaneous sympathetic vasodilatation is released from the hypothalamus. The vasodilatation implies a rise in bloodflow, whereby the rise in shell conductance from 20 to 90 W / oC is explained. In this phase the patient is red and warm.