Use the equation:
#q=mcDeltat#
where #q# is the heat gained or lost in Joules, #c# is specific heat capacity, #m# is mass, and #Deltat# is the temperature change, which is #(T_"final"-T_"initial")#.
The specific heat of water, #(c_"H2O")# is #4.184 "J"/("g"*^@"C")#
https://water.usgs.gov/edu/heat-capacity.html
Organize the Information
Given/Known
#c_"H2O"=4.184 "J"/("g"*^@"C")#
#m="750 g"#
#q="21966 J"#
Unknown: #Deltat#
Solution
Rearrange the equation to isolate #Deltat#. Substitute the given/known values into the equation and solve.
#Deltat=q/(mxxc)#
#Deltat=(21966color(red)cancel(color(black)("J")))/(750color(red)cancel(color(black)("g"))xx(4.184 color(red)cancel(color(black)("J"))/(color(red)cancel(color(black)("g"))*^@"C")))="7.0"^@"C"#
