Question

3. Calculate the energy required to convert 40g of ice at -20°C to 140°C steam at a pressure of 600kPa. Explain each phases

Answer 1

Ans :-

Step.1 :-Heating of Ice to 0 ⁰C

Heat required (Q₁) = Mass (m) x Specific heat capacity (C) x Change in temperature (ΔT)

= (0.040 kg).(2090 J/kg.⁰C).(20 ⁰C)

= 1672 J

Step.2 :- Melting of Ice

Q₂ = (0.040 kg).(3.33 x 10⁵ J/kg)

= 13320 J

Step.3 :-Heating of water at 100 ⁰C

Q₃= (0.040 kg).(4186 J/kg.⁰C).(100 ⁰C)

= 16744 J

Step.4 :- Convert water to steam

Q₄= (0.040 kg).(2.26 x 10⁶ J/kg.)

= 90400 J

Step.5 :- Heating of steam upto 140 ⁰C

Q₅= (0.040 kg).(2010 J/kg.⁰C).(140-100)⁰C

= 3216 J

So,

Heat required = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

= 1672 J + 13320 J + 16744 J + 90400 J + 3216 J

= 125352 J

Hence, Heat required = 125352 J