Answer to Question #213241 in Molecular Physics | Thermodynamics for Tonnoy

Question

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

Answer 1

Gives

m=200kg

$L_{ic}=335kJkg^{-1}$

$L_{s}=2260KJkg^{-1}$

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

0° ice to 0°water

Q₁=mL

$Q_1=200\times335=67000KJ$

0° water to100° water

$$Q_2=ms∆T=200\times4200\times100=84000KJ$$

100° water to100° steam

$Q_3=mL$

$Q_3=200\times2260=452000kJ$

$$Q=Q_1+Q_2+Q_3=67000+84000+452000=603000KJ=603MJ$$

option (c) is correct option