Answer
3.62°C
Work Step by Step
Given
Mass of ice cube= 14.7g
Mass of water= 324 g.
The amount of heat required for melting 1 mol of ice is 6.02 kJ.
To melt 8.5 g of ice is:
q= mC$\Delta t$.
q= ((14.7g x(1mol / 18.02 g )x(6.0 kJ/mol )x($10^3$J/ 1 kJ))
q= 4.91 x $10^3$ J
The amount of heat produced by water is calculated using:
q= mC$\Delta t$.
here m= 324g
C= 4.184 J/g
$\Delta t$= ?
$\Delta t$ = q/mC= ((4.91x $10^3$ J)/ (324g)x(4.184 J/g))
= 3.62°C
Change in temperature is 3.62°C.
