A total of 1500 calories of heat are added to 100.0 g of water initially at 10.0°C. What is the final temperature of the water?

Assume that there is no heat gain from and no heat loss to the surroundings.

q = +1500 cal
m = 100.0 g
c = 1 cal/(g °C)
ΔT = ? °C

q = m c ΔT
+1500 = 100.0 × 1 × ΔT
ΔT = +15.0°C

Final temperature = (10.0 + 15.0)°C = 25.0°C

When heat is added to water it is heated up and its heat level goes up which is measured by its temperature rise provided all the heat is getting absorbed by water and not lost to surrounding , thus thermodynamically water is to be considered as an isolated closed system.
Heat added Q cal = 1500 cal = weight of water ( 100 g ) x specific heat Cp of water ( 1 cal /g c ) x temperature difference ( final temp Tf - initial temp Ti = 10 C ) .
so , 1500 cal = 100 g x 1 cal / gC X Temperature difference ,
therefore , Temp. difference = (1500 cal) / 1g/C X 100 g = 15 C ,
so final temp of water = initial temp 10 C + 15 C = 25 C