J.R. S. answered 03/26/21
Ph.D. University Professor with 10+ years Tutoring Experience
They tell you to use the equation q = mc∆T. So, let's do that.
q = heat = ?
m = mass = 225 g
c = specific heat of water = 1.00 cal/gº
∆T = change in temperature = 43.7º - 10.5º = 33.2º
Plug in the values and solve for q
q = (225 g)(1.00 cal/gº)(33.2º) = 7470 calories