Principe noir – problèmes et solutions

Principe noir – problèmes et solutions

1. 200 grammes d'eau à 30 °C mélangés à 100 grammes d'eau à 90 °C. chaleur spécifique 1 cal/g d'eau = 1 cal/g^-1° C^-1Déterminez la température finale du mélange !

Connu :

Masse de l'eau 2 (m₁) = 200 grammes

Température de l'eau 1 (T₁) = 30^oC

Masse d'eau 2 (m₂) = 100 grammes

Température de l'eau 2 (T₂) = 90^oC

The specific heat for water (c) = 1 cal.gram^-1° C^-1

Recherché : La température finale

Solution:

La chaleur released by hooter water (Q₂) = heat absorbed by cooler water (Q₁)

m₂ c (ΔT) = m₁ c (ΔT)

(100)(1)(90 – T) = (200)(1)(T – 30)

(100)(90 – T) = (200)(T – 30)

9000 – 100T = 200T – 6000

9000 + 6000 = 200T + 100T

15000 = 300T

T = 15000/300

T = 50^oC

2. 60-gram at 90^oC mixed with 40-gram water at 25^oC. What is the final temperature of the mixture. The specific heat for water = 1 cal.g^-1.^oC^-1.

Connu :

Masse d'eau 1 (m₁) = 60 grammes

Température de l'eau 1 (T₁) = 90^oC

Masse d'eau 2 (m₂) = 40 grammes

Température de l'eau 2 (T₂) = 25^oC

The specific heat for water (c) = 1 cal.g^-1.^oC^-1

Recherché : La température finale

Solution:

Heat released by hooter water (Q₂) = heat absorbed by cooler water (Q₁)

m₁ c (ΔT) = m₂ c (ΔT)

(60)(1)(90 – T) = (40)(1)(T – 25)

(60)(90 – T) = (40)(T – 25)

5400 – 60T = 40T – 1000

5400 + 1000 = 40T + 60T

6400 = 100T

T = 6400/100

T = 64^oC

3. M-gram ice at 0^oC placé dans 340 grammes d'eau à 20^oC in a specific container. Latent heat of fusion (L_ourricier) = 80 cal g^-1, the specific heat of water (c_eau) = 1 cal g^{-1 o}C^-1. All the ice melts and the thermal equilibrium = 5^oC. Find mass of ice.

Connu :

Masse d'eau (m) = 340 grammes

Température de la glace (T_ourricier) = 0^oC

Température de l'eau (T_eau) = 20^oC

Temperature of thermal equilibrium (T) = 5^oC

Chaleur latente de fusion de la glace (L_ourricier) = 80 cal g^-1

The specific heat for water (c_eau) = 1 cal g^{-1 o}C^-1

Recherché : Mass of ice (M)

Solution:

Heat released by water (Q₂) = heat absorbed by ice (Q₁)

m_eau c_eau (ΔT) = m_ourricier L_ourricier +m_ourricier c_eau (ΔT)

(340)(1)(20-5) = M (80) + M (1)(5-0)

(340)(15) = 80M + 5M

5100 = 85 millions

H = 5100/85

M = 60 grammes

4. A copper at 100^oC placed in 128-gram of water at 30 ^oC. The specific heat for water is 1 cal.g^-1oC^-1 and the specific heat for copper is 0.1 cal.g^-1oC^-1. If the temperature of the thermal equilibrium = 36 ^oC, what is the mass of copper.

Connu :

Temperature of copper (T₁) = 100 ^oC

The specific heat for copper (c₁) = 0.1 cal.g^-1oC^-1

Masse d'eau (m₂) = 128 grammes

Température de l'eau (T₂) = 30 ^oC

The specific heat for water (c₂) = 1 cal.g^-1oC^-1

The temperature of the thermal equilibrium (T) = 36 ^oC

Recherché : Mass of copper (m₁)

Solution:

Q copper = Q water

m₁ c₁ ΔT = m₂ c₂ ΔT

(m₁)(0.1)(100-36) = (128)(1)(36-30)

(m₁)(0.1)(64) = (128)(1)(6)

(m₁)(6.4) = 768

m₁ = 768 / 6.4

m₁ = 120 gramme

5. A 3-kg lead with the specific heat for lead is 1400 J.kg^-1C^-1 à 80^oC placed in 10-kg water with the specific heat for water is 4200 J.kg^-1C^-1. The temperature of the thermal equilibrium is 20^oC. What is the initial temperature of water.

Connu :

Mass of lead (m₁) = 3 kg

The specific heat for lead (c₁) = 1400 J.kg^-1C^-1

Temperature of lead (T₁) = 80 ^oC

Masse d'eau (m₂) = 10 kg

La chaleur spécifique de l'eau (c₂) = 4200 J.kg^-1C^-1

The temperature of the thermal equilibrium (T) = 20 ^oC

Recherché : The initial temperature of water (T₂)

Solution:

Q libéré = Q absorbé

Q lead = Q water

m₁ c₁ ΔT = m₂ c₂ ΔT

(3)(1400)(80-20) = (10)(4200)(20-T)

(4 200)(60) = (42 000)(20-T)

252 000 = 840 000 – 42 000 T

42 000 T = 840 000 – 252 000

42,000 588,000 T = XNUMX XNUMX

T = 588,000 42,000 / XNUMX XNUMX

T = 14

The initial temperature of water is 14^oC.

6. 75-gram water at 0^oC mixed with 50-gram water so the temperature of mixture is 40^oC. What is the initial temperature of 50-gram water.

Connu :

Masse d'eau 1 (m₁) = 75 grammes

The initial temperature of water 1 (T₁) = 0^oC

Masse d'eau 2 (m₂) = 50 grammes

The temperature of mixture (T) = 40^oC

Recherché : The initial temperature of water 2 (T₂)

Solution:

Heat released by hooter water (Q_libérer) = heat absorbed by cooler water (Q_absorbe)

m₁ c (ΔT₁) = m₂ c (ΔT₂)

m₁ (ΔT₁) = m₂ (ΔT₂)

(75)(40 – 0) = (50)(T₂ - 40)

(75)(40) = (50)(T₂ - 40)

3000 = 50 T₂ au 2000 Février

3000 + 2000 = 50 T₂

5000 = 50 T₂

T₂ = 100 ^oC

7. 200-gram metal heated to 120^oC, then placed in 100-gram water at 30^oC. The temperature of mixture at thermal equilibrium is 60^oC. If the specific heat of water is 4200 J.kg^{-1 o}C^-1, what is the specific heat of the metal.

Solution:

Convert mass unit from gram to kilogram (International unit)

Connu :

Masse du métal (m₁) = 200 grammes = 0.2 kg

Temperature of metal (T₁) = 120^oC

Masse d'eau (m₂) = 100 grammes = 0.1 kg

Température de l'eau (T₂) = 30^oC

The temperature of mixture (T) = 60^oC

La chaleur spécifique de l'eau (c₂) = 4200 J.kg^{-1 o}C^-1

Recherché : The specific heat of metal (c₁)

Solution:

Q_libérer = Q_absorber

m₁ c₁ (ΔT₁) = m₂ c₂ (ΔT₂)

(0.2)(c)₁)(120 – 60) = (0.1)(4200)(60 – 30)

(0.2)(c)₁)(60) = (0.1)(4200)(30)

12 c₁ = 12600

c₁ = 12600 / 12

c₁ = 1050

1. Question: Who was Joseph Black? Réponse: Joseph Black was an 18th-century Scottish physician and chemist, known for his discoveries of latent heat, specific heat, and carbon dioxide.

2. Question: Qu'est-ce que la chaleur latente ? Réponse: Latent heat is the amount of energy absorbed or released by a substance during a phase change, like melting or boiling, without changing its temperature.

3. Question: How does latent heat differ from sensible heat? Réponse: While sensible heat causes a change in temperature of a substance without a phase change, latent heat causes a phase change without a change in temperature.

4. Question: What significance did Black’s experiments on latent heat have on the study of thermodynamics? Réponse: Black’s work laid the foundational understanding of heat energy and its conservation, paving the way for further advancements in thermodynamics and calorimetry.

5. Question: Why does ice at 0°C and water at 0°C have different amounts of heat energy? Réponse: Ice at 0°C needs to absorb the latent heat of fusion to become water at 0°C. Thus, water at 0°C has more heat energy than ice at the same temperature due to this absorbed latent heat.

6. Question: What is specific heat? Réponse: Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin.

7. Question: How did Black’s work on specific heat lead to the understanding of different heat capacities of substances? Réponse: Black discovered that equal masses of different substances required different amounts of heat to achieve the same change in temperature. This observation introduced the concept of specific heat capacities of substances.

8. Question: Why is the concept of latent heat crucial in weather systems? Réponse: Latent heat plays a significant role in weather systems because of the energy involved in the phase changes of water, like evaporation and condensation. This energy transfer affects temperature, cloud formation, and weather patterns.

9. Question: What apparatus did Joseph Black use to measure heat changes in his experiments? Réponse: Black used a simple calorimeter, an instrument for measuring heat change, in his experiments to quantify the amount of heat absorbed or released.

10. Question: How did Black’s discovery of carbon dioxide (which he called “fixed air”) contribute to chemistry? Réponse: Black’s identification of carbon dioxide as a distinct gas different from atmospheric air was pivotal in the advancement of gas chemistry and paved the way for understanding chemical reactions involving gases.