Question

What will be the final temperature if 1,68,000 J of heat is absorbed by 2 kg of water at 30°C?

Answer 1

Heat absorbed Q = 1,68,000 J
m = 2kg
Initial temperature t₁ = 30°C
Let final temperature t₂ = x°C
Rise is temperature Δt = (t₂ – t₁)
= (x – 30)°C
Specific heat capacity of water C = 4200 J kg^-1°C^-1
We know that Q = m × c × Δt
1,68,000 = 2 × 4200 × (x – 30 )
x – 30 = \(\frac{1,68,000}{2 \times 4200}\)
x- 30 = 20
x = 30 + 20
= 50°C
So, the final temperature of water = 50°C