Answer
1.1x $10^2$ g
Work Step by Step
Given
Volume of water= 352 mL.
density of water = 1.g /mL
Mass of water = volume x density
= 352 mL x 1 g/mL= 352 g.
The amount of heat produced by water is calculated using:
q= mC$\Delta t$.
here m= 352g
C= 4.184 J/g
$\Delta t$ = (0 C-25C)= -25 C
q= mC$\Delta t$ = (352 g)x( 4.184 J/g)x(-25C)
q= - 36819 J= -36.8 kJ
The amount of heat required for melting 1 mol of ice =6.02 kJ.
So, with -37 kJ
= 37kJx( 1mol/ 6.02 kJ)x( 18.02 g/ 1 mol h2O)
= 1.1 x $10^2$ g.
So with -37 kJ, we can melt 1.1x $10^2$ g of ice.
