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"#