Question

How much heat does it take to convert 0.500 kg of ice at -20^oC to steam at 250.^oC?

Heat capacities: ice, 2.1 J g^-1oC; liquid 4.2 J g^{-1 o}C^-1; steam 2.0 g^{-1 o}C^-1

Heat of fusion: 6.01 kJ/mol. Heat of vaporization: 40.7 kJ/mol.

Answer 1

Given that, m is the mass of ice = 0.500 kg = 500 g

First of all, Heat given will raise the temperature of ice from -20°C to 0°C.

Heat required for this process (q₁) = m*c*T

where m is the mass, c is the heat capacity of ice and T is the change in temperature

q₁ = (500 g) *(2.1 J/ g^oC) * (20^oC) = 21000 J

Now, heat is required to change the state of ice to liquid water

q₂ = n*H_fus

n is the number of mole of ice

q₂ = (500/18.02) * 6.01 kJ/mol = 166759.2 J

heat is required to raise the temperature of liquid water from 0°C to 100°C.

q₃ =  (500 g) *(4.2 J/ g^oC) * (100^oC) = 210000 J

Now, heat is required to change the state from liquid water to steam

q₄ = (500/18.02) * 40.7 kJ/mol = 1129301 J

Heat is required to raise the temperature of steam from 100°C to 250°C.

q₅ = (2.0 J/g°C)(500 g)(150°C) = 150000 J

So, total heat required = q₁ + q₂ + q₃ + q₄ + q₅ =  21000 J + 166759.2 J + 210000 J + 1129301 J +150000 J = 1677060.2 J = 1677.06 kJ