Question

Before going in for his annual physical, a 70.0 kg man whose body temperature is 37.0∘C consumes an entire 0.355 L can of a soft drink (mostly water) at 13.0∘C. His body releases energy at a rate of 7.00×103kJ/day (the basal metabolic rate, or BMR). Suppose that the specific heat of the man's body is 3480 J/kg⋅K. 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∘C? Assume that all of the released energy goes into raising the temperature. (mins)

Answer 1

Suppose Tf is the final temperature of the man's body.

Now, 0.355 litre of drink (mostly water) has a mass of 0.355kg

Specific heat capacity of water = 4187 J/KgK

Energy (mcθ) lost by body = energy gained by drink

70 x 3480 x (37-Tf) = 0.355 x 4187 x (Tf -13)

=> 9013200 - 243600Tf = 1486.4 Tf - 19323

=> 245086.4 *Tf = 9032523

=> Tf = 36.85 deg C.

Now,

Energy output = (7.0 x 10^3) J / 24 hours

Energy to be recovered = m x s x

= 70 x 3480 x (37 - 36.85) = 36920 J

Therefore, time needed to take the man's metabolism to return the temperature of his body

= 36920 x 24 / (7.0 x 10^3)

= 0.1266 hr = 7.59 Minutes (Answer)