Question

Calculate the energy absorbed when 13 kg of liquid water raises from 18°C to 100°C and then boils at 100°C.

Answer 1

Answer:

the energy absorbed is 4.477 x 10⁶ J

Explanation:

mass of the liquid, m = 13 kg

initial temperature of the liquid, t₁ = 18 ⁰C

final temperature of the liquid, t₂ = 100 ⁰C

specific heat capacity of water, c = 4,200 J/kg⁰C

The energy absorbed is calculated as;

H = mcΔt

H = mc(t₂ - t₁)

H = 13 x 4,200(100 - 18)

H = 4.477 x 10⁶ J

Therefore, the energy absorbed is 4.477 x 10⁶ J