Question

How much heat is required to melt 1 kg of solid water at 0 °C?

How much heat is released when freezing 1 kg of solid water at 0 °C?

How much heat is released when freezing 1 kg of supercooled water at -10 °C?

How much heat is required to boil 1 kg of superheated water at 110 °C?

How much heat is required to convert 1 kg of ice at -20 °C to water vapor at 120 °C?

Why could skin contact with 0.1 mol water vapor at 100 °C provide a more severe burn than 0.1 mol liquid water at 100 °C?

Answer 1

Solution :-

1) How much heat is required to melt 1 kg of solid water at 0 °C?

Solution :- heat of fusion of ice is 6.02 kJ/mol

Therefore

Amount of heat required to melt 1 kg ice is

q= (1 kg * 1000 g/1kg)*(1 mol / 18.0148 g per mol ) *(6.02 kJ/mol) = 334 kJ

so it will take 334 kJ of heat.

2) How much heat is released when freezing 1 kg of solid water at 0 °C?

Solution :-

Freezing is the opposite process of the melting which takes place at the constant temperature

Therefore the amount of heat realesed in freezing of 1 kg ice is = 334 kJ

We can also write it as -334 kJ because process is exothermic.

3) How much heat is released when freezing 1 kg of supercooled water at -10 °C?

Solution :- specific heat of ice is 2.03 J/g C

q = m*c*delta T + delta H fusion

= (1000 g * 0.00203 kJ per g C * 10 C) + 334 kJ

= 354.3 kJ

So it can release 354.3 kJ heat.

4) How much heat is required to boil 1 kg of superheated water at 110 °C?

Solution :- specific heat of steam is 0.00209 kJ/g C

Heat of vaporization of water 2.259 kJ/g

q = delta H vap + (m*c*deltaT)

= (2.259 kJ per g * 1000 g) + (1000 g * 0.00209 kJ per g C *10 C)

= 2259 kJ + 20.9 kJ

= 2280 kJ

5) How much heat is required to convert 1 kg of ice at -20 °C to water vapor at 120 °C?

Solution : -

q = (m*c*delta T solid)+ delta H fus + (m*c*delta T liq )+ delta H vap + (m*c*delta T steam)

q= (1000g *0.00203 *20)+(0.334 kJ /g * 1000 g) +(1000 g * 0.004184 kJ/g*100C) +(2.259 kJ/g * 1000g) +(1000 g * 0.00209 kJ/g* 20)

q= 3094 kJ

So it will take 3094 kJ heat

6) Why could skin contact with 0.1 mol water vapor at 100 °C provide a more severe burn than 0.1 mol liquid water at 100 °C?

Solution :-

Liquid water can have 100 C as the maximum temperature but the steam can have higher temperature than liquid water

Therefore when same mol water and steam is contacts with skin then steam can cause more burn than liquid water.