Chapter 18 - Temperature, Heat, and the First Law of Thermodynamics - Problems - Page 542: 25

$V=94.6 Liter$

As we know that 1 food calorie= 1000 cal The required mass of water that needs to drink can be calculated as follows: $m=\frac{Q}{c\Delta{T}}$ putting the values, we get $m=\frac{3500\times1000(cal)}{1(cal/g.K)\times(37-0)C^{\circ}}=94.6\times10^3g$ $V=\frac{m}{\rho}$ putting the values, we get $V=\frac{94.6\times10^3 g}{1000g/Liter}=94.6 Liter$