Question #211554

Heat is added to 200 kg of ice at 0 ºC until dry steam at 100 ºC is obtained.

The specific heat capacity of water is 4200 J kg‑1 K‑1, the specific latent heat of steam is 2260 kJ kg-1 and the specific latent heat of ice is 335 kJ kg-1.

Determine the total amount of heat energy (in MJ) added to the ice.


1
Expert's answer
2021-06-29T08:22:10-0400

Gives

M=200kg

sw=4200jkg1K1,s_w=4200jkg^{-1}K^{-1},

Letent heat L=2260kj/kgL=2260kj/kg

sice=335kJ/kgs_{ice}=335kJ/kg

0° ice to 0° water

Q=mLwQ=mL_w


Q1=200×4200=840000KJQ_1=200\times4200=840000KJ

0° water to 100°water

Q=mswTQ=ms_w∆T


Q2=200×335×(1000)=670000KJQ_2=200\times335 \times(100-0)=670000KJ

100° water to 100steam


Q3=msteamL=200×2260=452000KJQ_3=m_{steam}L=200\times2260=452000KJ

Qt=Q1+Q2+Q3=840000+670000+452000=1.962MJQ_t=Q_1+Q_2+Q_3=840000+670000+452000=1.962MJ


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