Answer:
ΔT = 28.667°C is the answer.
Explanation:
Given, Mass of water = 4.0 kg
Water is heated by = 8 × 10² W = 800 W
Time = 10 min = 10 × 60 = 600 s
Now, using the equation of specific heat,
Q = mcΔT
where c is the specific heat capacity of water = 4186 J/kg°C
Q = Pt
Substituting,
Pt = mcΔT
800 × 600 = 4 × 4186 × ΔT
480000 = 16744 × ΔT
ΔT = 28.667
Temperature increased, ΔT = 28.667°C
