A large pot is placed on a stove and 1.2 kg of water at 14°C is added to the pot. The temperature of the water is raised evenly to 100°C just before it starts to boil. (a) What amount of heat is absorbed by the water in reaching 100°C? (b) The water then boils until all of it has evaporated, turning to water vapor at 100°C. How much heat does the water absorb in this process?

a) the amount of heat absorbed by the water is 431995.2 J

b) the amount of heat absorbed during evaporation is 2712000 J

Explanation:

Given that;

mass of water Mw = 1.2 kg

Specific heat capacity of water Cw = 4186 J/kg.C

Change in temperature ΔT = final T - Initial T = 100 - 14 = 86°C

Now

A)

Heat required to raise the temperature of water is expressed as:

Q = Mw × Cw × ΔT

Q = 1.2 × 4186  × 86

Q = 431995.2 J

Therefore the amount of heat absorbed by the water is 431995.2 J

B)

Then heat absorbed during evaporation will be:

Q1 = Heat absorbed during phase change from water to steam = Mw × Lv

Lv = latent heat of vaporization of water = 2.26 × 10⁶ J/kg

so

Q1 = 1.2 × 2.26 × 10⁶ = 2712000 J

Therefore the amount of heat absorbed during evaporation is 2712000 J

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

d eras and periods