Question

A 60 kg boy running at `5.0 m//s` while playing basketball falls down on the floor and skids along on his leg until he stopes. How many calories of heat are generated between his leg and the floor?
Assume that all this heat energy is confined to a volume of `2.0cm^3` of his flesh. What will be temperature change of the flesh? Assume `c=1.0(cal)/(g^@C)` and `rho=950(kg)/(m^3)` for flesh.

Answer 1

Kinetic energy of boy `K=(1)/(2)mv^2=(1)/(2)(60)(5)^2=750J`
Changing this energy in joule into calorie
`Q_1=(K)/(J)=(750)/(4.2)=179J` .(i)
As this kinetic energy changes into heat energy `(Q_2)` and this heat is confined to flesh of boy,
`Q_2=mcDeltaT=(rhoV)cDeltaT` .(ii)
As `Q_1=Q_2`, hence from Eqs. (i) and (ii)
`179=0.95xx2xx2xxDeltaTimpliesDeltaT=94^@C`