Summary of "PERUBAHAN WUJUD KALOR & ASAS BLACK | IPA Kelas 7 SMP BAB 5"

Main ideas and lessons

Heat changes during state transitions can be represented on a graph:
- Some segments show temperature change (heating/cooling while the substance remains in the same phase).
- Other segments show phase change (e.g., melting or evaporation) where heat is added without an immediate temperature change.
The amount of heat depends on whether:
- The substance’s temperature changes (use specific heat).
- The substance undergoes melting/evaporation (use latent heat).
When hot and cold substances are mixed, thermal equilibrium occurs:
- Heat released = heat received — this is the core of Black’s principle.
- The final temperature is found by writing an equation that sets both sides equal.

For segments where temperature changes (e.g., labeled Q1, Q3, Q5):
- Use: [ Q = m \cdot c \cdot \Delta t ] where:
  - (Q) = heat (J)
  - (m) = mass (kg)
  - (c) = specific heat (J/kg°C)
  - (\Delta t) = temperature change (°C)
For a melting segment (e.g., labeled Q2):
- Use: [ Q = m \cdot l ] where:
  - (l) = heat of fusion/melting (J/kg)
For an evaporation (vaporization) segment (e.g., labeled Q4):
- Use: [ Q = m \cdot u ] where:
  - (u) = heat of steam/vaporization (J/kg)
Total heat for the full path on the graph:
- [ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 ]
- (Expressed in the subtitles as a sum of the segment heats.)

Given:

Steps:

Process P → Q (temperature change in ice):
- [ Q_1 = m \cdot c \cdot \Delta t ]
- Substituting (as stated): [ Q_1 = 1 \cdot 2100 \cdot 0.5 = 10{,}500\ \text{J} ]
Process Q → R (melting):
- [ Q_2 = m \cdot l ]
- Substituting: [ Q_2 = 1 \cdot 336{,}000 = 336{,}000\ \text{J} ]
Total for Q → R (concluded in the subtitles):
- [ Q_{\text{PQR}} = Q_1 + Q_2 = 10{,}500 + 336{,}000 = 346{,}500\ \text{J} ]

Core concept:

When mixing two materials with different initial temperatures:
- Heat released by the hotter object = heat received by the colder object.
Thermal equilibrium is the state where both reach the same final temperature.

Equation format (as given): [ M_1 \cdot C_1 \cdot (T_1 - t_a) = M_2 \cdot C_2 \cdot (t_a - T_2) ] where:

(M_1, C_1, T_1): mass, specific heat, and initial temperature of the hotter substance
(M_2, C_2, T_2): mass, specific heat, and initial temperature of the colder substance
(t_a): final equilibrium temperature

Given (as stated):

Cold water:
- (m_{\text{cold}} = 200\ \text{g})
- (T_{\text{cold}} = 20^\circ\text{C})
Hot water:
- (m_{\text{hot}} = 100\ \text{g})
- (T_{\text{hot}} = 80^\circ\text{C})
Specific heat of water (as used in the subtitle): 1 kilocalorie/(g·°C)

Set heat lost = heat gained:

Hot side heat released (as written):
- [ 100 \cdot 1 \cdot (80 - t_a) ]
Cold side heat received:
- [ 200 \cdot 1 \cdot (t_a - 20) ]
- (The subtitle’s algebra formatting is slightly off, but it leads to the same final result.)