Answer
(a) $0.38J$
(b) $0.54mK$
Work Step by Step
(a) The energy absorbed can be determined as
$E=(Dose \space in\space rad)m=225rad(0.17Kg)(\frac{0.01J/Kg}{1rad})78Kg$
$E=0.38J$
(b) We can find the increase in temperature as
$\Delta T=\frac{Q}{mc}$
We plug in the known values to obtain:
$\Delta T=\frac{0.3825J}{(0.17Kg)(4186J/KgK)}$
$\Delta T=0.54mK$
