Answer
$\Delta T_{body}=-0.6°C$
Work Step by Step
The mass of 1L of water $\rho=1kg/L$ is 1kg.
The energy balance for the system:
$\Delta U_{body}+\Delta U_{water}=0$
Given $m_{body}=68\ kg,\ c_{v,body}=3.6\ kJ/kg.K,\ c_{v,water}=4.18\ kJ/kg.K$
$T_{1,body}=39°C,\ T_{1,water}=5°C$:
and solving for the final temperature: $T_2=38.4°C$
Hence: $\Delta T_{body}=-0.6°C$
