Question

if 12.552 joules of heat is applied to 3.0 g of liquid water, what is the change in temperature? the specific heat of liquid water is 4.184 j g-1 °c-1.

Answer 1

Q = mc(delta T)
12.552 = (3.0)(4.184)(delta T)
Delta T = 1

Hence, Temperature increases by 1 degree Celsius.