Answer:
10 degree C
Explanation:
Q = 500 kcal = 500 x 1000 x 4.186 J = 2.1 x 10^6 J
V = 50 liter
m = Volume x density = 50 x 10^3 x 1000 = 50 kg
Let ΔT be the rise in temperature.
Specific heat of water = 4186 J/kg C
Q = m x c x ΔT
2.1 x 10^6 = 50 x 4186 x ΔT
ΔT = 10 degree C