Please check my answer of chapter heat

Dear Student,
Given:
Mass, m = 6 kg 
T₁ = 30°C           T₂ = 60°C
∆T = T₂ - T₁ = 60° - 30° => 30°C 
Specific heat capacity of water, C = 4200 J/kg°C
We know that,
Q = m × C × ∆T
Q = 6 × 4200 × 30
Q = 756000 J
Since, 1 Joule = 2.3 × 10^-4 cal
756000 J = 17388 × 10^-2 cal => 173.88 cal

Regards