Answer:
The question requires the value of the initial temperature which is found to be 60.266 °C ≅ 60.3 °C
Explanation:
To solve this question we list out the given variables as follows
Mass of water = 3110 g
Heat energy absorbed = 5.19 × 10⁵ J
Heat Energy required to raise the temperature of water to boiling point =
ΔH = m×C×Δθ where
m = mass of waster = 3110 g = 3.11 kg
C = specific heat capacity of water = 4.2 J/g°C
Δθ = Temperature change = T₂ - T₁ and
T₂ = 100°C which is the normal boiling point temperature of water
Hence 5.19 ×10⁵ J = 3110 g × 4.2 J/g°C ×Δθ
from where Δθ = (5.19 ×10⁵ J)/(13062 J/°C) = 39.73 °C
But Δθ = T₂ - T₁ = 100 °C- T₁ = 39.73 °C then
T₁ = -100 °C - 39.73 °C = 60.266 °C
Hence the initial temperature of the water is 60.266 °C
