Before going in for his annual physical, a 70.0 kg man whose body temperature is

Question

Before going in for his annual physical, a 70.0 kg man whose body temperature is 37.0 degree C consumes an entire 0.355 L can of a soft drink (mostly water) at 15.0 degree C. His body releases energy at a rate of 7.00 times 10^3 kJ/day (the basal metabolic rate, or BMR). How long will this take the man's metabolism to return the temperature of his body (and of the soft drink that he consumed) to 37.0 degree C.? Assume that all of the released energy goes into raising the temperature. T= 7.52 min

Explanation / Answer

Assuming that body and water have equal specific heats,
Tm = (m1*T1 + m2*T2)/(m1+m2)
Tm = (70*37 + 0.355*15) / (70.355) = 36.88 °C.

the heat needed to heat the body to 37° is = m*c*T
Q = 4.18 * 70.355 * 0.111 kJ
Q = 32.64 kJ

Heat production is (7*10^3 kJ)/(24*60) min = 4.86 kJ/min
time needed to produce  32.64 kJ , =  32.64/4.86 min = 6.7145 min

t = 6.71 min

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