Question

How much water will a man of 80 kg drink to maintain body temp is 45${}^0$C in a day.

• A. 0.7 L
• B. 0.69 L
• C. 0.9 L
• D. 1.21 L

Answer 1

Answer: (B)

0.69 L

Energy released due to change in temperature=energy required for phase change
$M_{canel}{C}_{canel}\Delta T=m_{water}L$
or $m_{water}=\frac{80\times 3480\times 6}{2.42\times {10}^6}=690247.9Kg$