States of matter

Solution:

Substances can exist in three states (phases) – solid, liquid, and gas.
Melting – change from solid to liquid
Freezing – change from liquid to solid
Evaporation – change from liquid to gas
Boiling – evaporation occurring when the vapor pressure equals the external pressure
Condensation – change from gas to liquid
Sublimation – direct change from solid to gas

                   
L = latent heat (melting, freezing, evaporation, boiling ...)
l = specific latent heat (melting, freezing, evaporation, boiling ...)
m = mass of substance
                   
For ice – water – steam applies:
t _melt = t _freeze = 0⁰C,      t _boil = t _condense = 100⁰C,    

l _melt = l _freeze = 334.10³J.kg^-1,    l _boil = l _condense = 2260.10³J.kg^-1

l _evap(20⁰C) = 2430.10³J.kg^-1,  l _sub = 2830.10³J.kg^-1

c(H₂O) = 4180 J.kg^-1.K^-1,   c(ice) = 2100 J.kg^-1.K^-1

Conversion of ice to steam:    Q = Q₁ + L₁ +Q₂ + L₂ + Q₃
Q₁ = heat needed to warm ice to melting point
L₁ = heat needed to convert ice to water
Q₂ = heat needed to warm water to boiling point
L₂ = heat needed to convert water to steam
Q₃ = heat needed to warm steam

Solution:

Analysis:

m = 10 kg,  t₁ = 20⁰C,  t_m (Al) = 660⁰C,  c(Al) = 896 J.kg^-1.K^-1
l_m (Al) = 400.10³J.kg^-1

To melt the aluminum object, approximately 9.73 MJ of heat is required.

Solution:

Analysis:

m = 5 kg,  L = 1.02.10⁶J,

The specific latent heat of fusion of copper is l_m (Cu) = 204,000 J.kg^-1

Solution:

Analysis:

m₁ = 5 l = 5 kg,  t₁ = 50⁰C,  Δ t = 50⁰C,  c1(H₂O) = 4180 J.kg^-1.K^-1
l_m (ice) = 334,000 J.kg^-1

The mass of ice must be m = 3.13 kg.

Solution:

Analysis:

m = 10 kg,  Δt = 100⁰C = 100 K,  c₁ = 4180 J.kg^-1.K^-1,
l_v = 2,26·10⁶ J.kg^-1

Heat needed to warm the water:
$Q_1 = 4180\,\text{kJ}$

$p_1=\frac{4180}{26780}=15.6\mathrm{\%}$

Heat needed for the phase change:
$L_v=22600\text{kJ}$

$p_2=\frac{22600}{26780}=84.4\mathrm{\%}$

The water received a total of 26,78 MJ of heat.

Solution:

Analysis:

m = 0,5 kg,  Δt = 970⁰C − 20⁰C = 950 K,  Q = 2,67·10⁵ J, c(brass) = 394 J.kg^-1K^-1

The specific latent heat of fusion of brass is l_t = 159,7 kJ.kg^-1

Solution:

Analysis:

t_t = 970⁰C,  c = 394 J.kg^-1.K^-1,  l_t = 159 700 J.kg^-1, Q = 441 980 J
m₁ = 1 kg,  m₂ = 0,5 kg,  t₁= ?

The initial temperature of the brass body was 50⁰C.

Solution:

Analysis:

H = 30·10⁶ J.kg^-1,  m₁ = 6·10³ kg,  t₁ = 10⁰C,  m₂ = 10³ kg,  t₂ = 100⁰C
c(water) = 4180 J.kg^-1.K^-1,  l_V = 2,26·10⁶ J.kg^-1,  η = 0,7

The mass of coal is 215 kg.

Solution:

Analysis:

t₁ = 27⁰C,   t_t = 327⁰C, l_t = 22 600 J.kg^-1,  c = 125 J.kg^-1K^-1
Δt = 327⁰C − 27⁰C = 300⁰C = 300 K

The speed of the lead ball must be v = 346,7 m.s^-1.

Solution:

ks – sublimation curve

kt – melting curve

kp – saturated steam curve

1) solid, ice

2) liquid, water

3) superheated steam

4) gas

Triple point: (T_A = 273,16 K, p_A = 610 Pa). At the temperature and pressure of the triple point, ice, water, and saturated steam can coexist in a closed vessel. Critical point: (T_K = 647,3 K, p_K = 22,13 MPa, ρ_K(H₂O) = 315 kg.m^-3). At the critical point, the “critical state of a substance” occurs. The difference between a liquid and its saturated vapor disappears. The substance becomes homogeneous. Saturated steam is in equilibrium with its condensed phase. Superheated steam is formed by further heating of saturated steam in the absence of liquid. It resembles an ideal gas. The equation of state applies. Gas is superheated steam heated to a temperature higher than the critical temperature.

Solution:

Analysis:

The system received 6,066 MJ of heat.

Solution: 

Analysis:

The mass of the condensed steam is 0,71 kg.

Solution:

Analysis:

The desublimation lasted 56,6 minutes.

Solution:

Analysis:

The initial temperature of the body was −7,9⁰C.

Solution:

Analysis:

The mass of the ice is 91 grams.

Solution:

Analysis:

The heating of the water lasted 45,3 minutes.

Solution:

Analysis:

The final temperature of the water is 47,65⁰C.

Solution:

Analysis:

We must put 63 ice cubes into the calorimeter.

Solution:

Analysis:

When the pond freezes, 307,3 GJ of energy is released.

Solution:

Analysis:

The internal energies of water and steam differ by 0,704 MJ.