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What are the answers to questions 8, 9, 10, and 11?

Physics

ScanSolve AI Answer

Answer

C

specific latent heat of fusion

Step-by-step solution

4 steps

here are the solutions to questions 8, 9, 10, and 11.

8. The quantity of heat energy needed to freeze one kilogram of milk at its freezing point is known as

When a substance freezes, it undergoes a phase change at a constant temperature, releasing heat. The heat energy required per unit mass for a phase change (like freezing or melting) is called the specific latent heat. For freezing or melting, it's specifically the specific latent heat of fusion.

The correct option is C. specific latent heat of fusion.

9. In a certain process, 12000 calories is absorbed as internal energy. (1 cal = 4.184J)

To convert calories to Joules, multiply the number of calories by the conversion factor.

  1. Convert calories to Joules.

    QJoules=Qcalories×conversion factorQ_{\text{Joules}} = Q_{\text{calories}} \times \text{conversion factor} QJoules=12000 cal×4.184JcalQ_{\text{Joules}} = 12000 \text{ cal} \times 4.184 \frac{\text{J}}{\text{cal}} QJoules=50208 JQ_{\text{Joules}} = 50208 \text{ J}

    The correct option is D. 50,208J.

    10. Some water is heated in a pot. The major mode(s) of heat transfer within the water is/are

    When water is heated in a pot, the water at the bottom heats up, becomes less dense, and rises. Cooler, denser water from the top sinks to take its place, gets heated, and rises. This continuous circulation of fluid is the primary way heat is transferred within the water.

    The correct option is A. convection.

    11. Calculate the quantity of heat required to completely convert 20g of ice at 0°C to water at the same temperature. (L_f = 336J/g)

    To calculate the heat required for a phase change (melting ice to water) at a constant temperature, use the formula:

    Q=mLfQ = mL_f

    where QQ is the heat energy, mm is the mass, and LfL_f is the specific latent heat of fusion.

  2. Identify the given values.

    m=20 gm = 20 \text{ g} Lf=336JgL_f = 336 \frac{\text{J}}{\text{g}}
  3. Calculate the heat energy in Joules.

    Q=(20 g)×(336Jg)Q = (20 \text{ g}) \times \left(336 \frac{\text{J}}{\text{g}}\right) Q=6720 JQ = 6720 \text{ J}
  4. Convert the heat energy to kilojoules (kJ).

    Q=6720 J1000JkJQ = \frac{6720 \text{ J}}{1000 \frac{\text{J}}{\text{kJ}}} Q=6.72 kJQ = 6.72 \text{ kJ}

    The closest option is D. 6.7KJ.

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