Question

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -12.0 ∘C, to steam at 109.0 ∘C?

Answer 1

Answer:

39.699 kJ

Explanation:

In this situation, there are a few transformations as follows;

(i) Heat required to warm the ice from -12°C to its melting point.

(ii) Heat required to melt the ice.

(iii) Heat required to boil the melted ice to boiling point (i.e to steam)

(iv) Heat required to vapourize the water

(v) Heat required to heat the steam from 100°C to 109.0°C

The sum of all the heat processes gives the heat required to warm the ice to steam;

Calculate each of these heat processes

From (i);

Let the heat required to warm the ice from -12.0°C to its melting point (0°C) be Q₁.

Q₁ = m x c x ΔT        -----------------------(i)

Where;

m = mass of ice = 13.0g

c = specific heat capacity of ice = 2.09 J/g°C

ΔT = final temperature - initial temperature = 0°C - (-12°C) = 12°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 2.09 x 12 = 326.04 J

From (ii);

Let the heat required to melt the ice be Q₂. This heat is called the heat of fusion and it is given by;

Q₂ = m x L        -----------------------(ii)

Where;

m = mass of ice = 13.0g

L = latent heat of fusion of ice = 333.6 J/g

Substitute these values into equation (ii) as follows;

Q₂ = 13.0 x 333.6

Q₂ = 4336.8 J

From (iii);

Let the heat required to boil the melted ice from 0°C to boiling point of 100°C be Q₃.

Q₃ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of melted ice (water) which is still 13.0g

c = specific heat capacity of melted ice (water) = 4.2 J/g°C

ΔT = final temperature - initial temperature = 100°C - 0°C = 100°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 4.2 x 100 = 5460 J

From (iv);

Let the heat required to vaporize the water (melted ice) be Q₄. This heat is called the heat of vaporization and it is given by;

Q₄ = m x L        -----------------------(iv)

Where;

m = mass of ice = 13.0g

L = latent heat of vaporization of water = 2257 J/g

Substitute these values into equation (iv) as follows;

Q₄ = 13.0 x 2257

Q₄ = 29341 J

From (v);

Let the heat required to heat the steam from 100°C to 109°C be Q₅.

Q₅ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of steam which is still 13.0g

c = specific heat capacity of steam = 2.01 J/g°C

ΔT = final temperature - initial temperature = 109.0°C - 100°C = 9°C

Substitute these values into equation (i) as follows;

Q₅ = 13.0 x 2.01 x 9 = 235.17J

Finally:

Sum all the heat values together;

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 326.04 + 4336.8 + 5460 + 29341 + 235.17

Q = 39699.01 J

Q = 39.699 kJ

Therefore, the amount of heat (in kJ) required is 39.699