During a long run a very well-trained dog can use up to 1000 ‘cal’/hour (note: food calories differ by a factor of one thousand from the scientific calorie). if it is assumed that the dog has a mass of twenty five kilograms and the physical properties of water, determine the uniform heat generation per unit volume under the assumption that the heat generation is uniform in the dog’s body(note: you must determine the dog’s volume as part of the

46488.8 W / m³

Explanation:

Given:

calories used by the dog = 1000 cal/ hour

1 Food calories = 1000 scientific calories

also,

1 scientific calorie = 4.184 J

thus,

total Power used by the dog per second = ( 1000 × 1000 ×  4.184 ) / 3600

or

total power used by the dog per second =  1162.22 W

Mass of the dog = 25 kg

since, the physical properties of the dog is similar to water

thus,

density of the dog = 1000 kg/m³

thus,

the volume of the dog = Mass / Density = 25 / 1000 = 0.025 m³

Now,

we have the relation

Heat supplied (Q) = volume × Heat generated per unit volume (q)

or

1162.22 = 0.025 × q

or

q = 46488.8 W / m³

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