Question

Water in the Antarctica usually exists as ice ,making it very difficult to use the water or conduct experiments reqiring the water to be in liquid state .A coffee addicted physisits stationed at a research base requires about 0.8 kg of ice for his coffe on the hour

Answer 1

Answer:

2730.304 KJ

Explanation:

 How much heat is required to convert 0.8 kg of ice at  -35°C into steam at 100 C?

Given that:

mass of ice (m) = 0.8 kg = 800 g

Initial temperature ($T_i$) = -35°C = 238 K

final temperature ($T_n$)= 100°C = 373 K

Specific heat of ice ($S_i$) = 2.108 J/g.K

Specific heat of water ($S_w$) = 4.18 J/g.K

Latent heat of fusion ($L_f$) = 334 J/g.

Latent heat of vaporization ($L_v$) = 2230 J/g.

$\Delta T=T_n-T_i=373-238=135K$

Total heat (Q) required to increase the temperature of ice from the initial temperature of 238K to final temperature of 373 K is given by the equation:

$Q=m\Delta TS_i+m\Delta TS_w+mL_f+mL_v\\Q=800(135)*2.108+800(135)*4.18+800*334+800*2230\\Q=227664+451440+267200+1784000=2730304J\\Q=2703.304KJ$