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Chapter 3: Q43E (page 576)

In the situation described in Exercise 17.42, the man’s metabolism will eventually return the temperature of his body (and of the soft drink that he consumed) to 37.00C. If his body releases energy at a rate of 7.00×103kJ/day(the basal metabolic rate, or BMR), how long does this take? Assume that all of the released energy goes into raising the temperature.

Short Answer

Expert verified

The metabolism will take7.6minto bring back the body to the initial temperature.

Step by step solution

01

Heat flow equation

The heat flow equation in a body of mass and specific heat capacity c for a temperature change ofδT is given by Q=mcδT.

02

Calculate the time using the heat flow equation

The rate of release of heat is 7×103kJ/day=81.01J/s.

Mass is , specific heat capacity is 3480J/kg0Cand change in temperature is0.150C .

Using the heat flow equation

ΔQΔt=mcδTtt=mcδTΔQΔtt=70×3480×0.1581.01t=456st=7.6min

So, the metabolism will take 7.6minto bring back the body to the initial temperature.

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