Question #213241

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.

a. 452MJ

b. 520MJ

c. 603MJ

d. 84MJ


1
Expert's answer
2021-07-06T11:37:05-0400

Gives

m=200kg

Lic=335kJkg1L_{ic}=335kJkg^{-1}

Ls=2260KJkg1L_{s}=2260KJkg^{-1}

s=4200Jkg1K1s=4200Jkg^{-1}K^{-1}

0° ice to 0°water

Q1=mL

Q1=200×335=67000KJQ_1=200\times335=67000KJ

0° water to100° water



Q2=msT=200×4200×100=84000KJQ_2=ms∆T=200\times4200\times100=84000KJ

100° water to100° steam

Q3=mLQ_3=mL

Q3=200×2260=452000kJQ_3=200\times2260=452000kJ


Q=Q1+Q2+Q3=67000+84000+452000=603000KJ=603MJQ=Q_1+Q_2+Q_3=67000+84000+452000=603000KJ=603MJ

option (c) is correct option


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