Answer to Question #211554 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.

Answer 1

Gives

M=200kg

$s_w=4200jkg^{-1}K^{-1},$

Letent heat $L=2260kj/kg$

$s_{ice}=335kJ/kg$

0° ice to 0° water

$Q=mL_w$

$$Q_1=200\times4200=840000KJ$$

0° water to 100°water

$Q=ms_w∆T$

$$Q_2=200\times335 \times(100-0)=670000KJ$$

100° water to 100steam

$$Q_3=m_{steam}L=200\times2260=452000KJ$$

$$Q_t=Q_1+Q_2+Q_3=840000+670000+452000=1.962MJ$$