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The provided text appears to be a solution set for a physics homework assignment. It contains answers and explanations for questions numbered 60, 61, and 62. Question 60 defines latent heat and asks f

Physics

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Answer

B

Change phase without temperature change

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Here are the solutions to the questions:

60. Latent heat is the heat required to:
Latent heat is the energy absorbed or released by a substance during a phase change (like melting, freezing, boiling, or condensation) at a constant temperature.

The correct option is B. Change phase without temperature change.

61. Which process involves latent heat?
Latent heat is involved in processes where a substance changes its physical state (phase).

  • A. Heating water from 20∘C20^\circ\text{C} to 50∘C50^\circ\text{C} involves a temperature change, not a phase change.
  • B. Boiling water at 100∘C100^\circ\text{C} involves a phase change from liquid water to steam (gas) at a constant temperature of 100∘C100^\circ\text{C}. This requires latent heat of vaporization.
  • C. Cooling metal involves a temperature change, not a phase change.
  • D. Compressing gas involves work done on the gas, which can change its temperature, but not necessarily a phase change requiring latent heat.

The correct option is B. Boiling water at 100∘C100^\circ\text{C}.

62. A 0.5 kg aluminum block (c = 900 J/kg°C) at 200°C is placed in 1 kg of water at 20°C. Final temperature is 40°C. Heat lost by aluminum is:

  1. Identify the given values for the aluminum block.

    Mass of aluminum, m=0.5 kgm = 0.5 \text{ kg}
    Specific heat capacity of aluminum, c=900 J/kg∘Cc = 900 \text{ J/kg}^\circ\text{C}
    Initial temperature of aluminum, Ti=200∘CT_i = 200^\circ\text{C}
    Final temperature of aluminum, Tf=40∘CT_f = 40^\circ\text{C}

  2. Calculate the change in temperature for the aluminum block.

    ΔT=Tf−Ti=40∘C−200∘C=−160∘C\Delta T = T_f - T_i = 40^\circ\text{C} - 200^\circ\text{C} = -160^\circ\text{C}

  3. Calculate the heat lost by the aluminum block using the formula Q=mcΔTQ = mc\Delta T.

    Since we are looking for heat lost, we take the absolute value.

    Q=∣mcΔT∣Q = |m c \Delta T| Q=∣0.5 kg×900 J/kg∘C×(−160∘C)∣Q = |0.5 \text{ kg} \times 900 \text{ J/kg}^\circ\text{C} \times (-160^\circ\text{C})| Q=0.5×900×160 JQ = 0.5 \times 900 \times 160 \text{ J} Q=450×160 JQ = 450 \times 160 \text{ J} Q=72,000 JQ = 72,000 \text{ J}

    The correct option is C. 72,000 J.

    63. Why is final temperature the same in calorimetry problems?
    In calorimetry problems, when objects at different temperatures are brought into contact within an isolated system, heat energy is exchanged between them until they reach a state of thermal equilibrium. At thermal equilibrium, all parts of the system have the same temperature.

    The correct option is B. Heat flows until thermal equilibrium.

    64. How much heat is required to convert 1 kg ice at 0°C to water at 100°C? (c = 4200 J/kg°C, L = 334,000 J/kg)

  4. Calculate the heat required to melt 1 kg of ice at 0∘C0^\circ\text{C} to 1 kg of water at 0∘C0^\circ\text{C} (latent heat of fusion).

    Given mass m=1 kgm = 1 \text{ kg} and latent heat of fusion L=334,000 J/kgL = 334,000 \text{ J/kg}.

    Q1=mLQ_1 = m L Q1=1 kg×334,000 J/kgQ_1 = 1 \text{ kg} \times 334,000 \text{ J/kg} Q1=334,000 JQ_1 = 334,000 \text{ J}

  5. Calculate the heat required to raise the temperature of 1 kg of water from 0∘C0^\circ\text{C} to 100∘C100^\circ\text{C}.

    Given mass m=1 kgm = 1 \text{ kg}, specific heat capacity of water c=4200 J/kg∘Cc = 4200 \text{ J/kg}^\circ\text{C}, and temperature change ΔT=100∘C−0∘C=100∘C\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100^\circ\text{C}.

    Q2=mcΔTQ_2 = m c \Delta T Q2=1 kg×4200 J/kg∘C×100∘CQ_2 = 1 \text{ kg} \times 4200 \text{ J/kg}^\circ\text{C} \times 100^\circ\text{C} Q2=420,000 JQ_2 = 420,000 \text{ J}

  6. Calculate the total heat required.

    Qtotal=Q1+Q2Q_{total} = Q_1 + Q_2 Qtotal=334,000 J+420,000 JQ_{total} = 334,000 \text{ J} + 420,000 \text{ J} Qtotal=754,000 JQ_{total} = 754,000 \text{ J}

    The correct option is B. 754,000 J. (Option C, 420,000 + 334,000 J, also represents the correct sum).

    65. A steel rod of 2 m expands by 1.2 mm when heated. Find temperature change. (α=1.2×10−5/∘C\alpha = 1.2 \times 10^{-5} /^{\circ}\text{C})

  7. Identify the given values and ensure consistent units.

    Original length, L0=2 mL_0 = 2 \text{ m}
    Change in length, ΔL=1.2 mm=1.2×10−3 m\Delta L = 1.2 \text{ mm} = 1.2 \times 10^{-3} \text{ m}
    Coefficient of linear expansion, α=1.2×10−5/∘C\alpha = 1.2 \times 10^{-5} /^\circ\text{C}

  8. Use the formula for linear thermal expansion, ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T, and rearrange to solve for ΔT\Delta T.

    ΔT=ΔLL0α\Delta T = \frac{\Delta L}{L_0 \alpha}

  9. Substitute the values and calculate ΔT\Delta T.

    ΔT=1.2×10−3 m2 m×1.2×10−5/∘C\Delta T = \frac{1.2 \times 10^{-3} \text{ m}}{2 \text{ m} \times 1.2 \times 10^{-5} /^\circ\text{C}} ΔT=1.2×10−32.4×10−5 ∘C\Delta T = \frac{1.2 \times 10^{-3}}{2.4 \times 10^{-5}} \text{ }^\circ\text{C} ΔT=1.22.4×10(−3−(−5)) ∘C\Delta T = \frac{1.2}{2.4} \times 10^{(-3 - (-5))} \text{ }^\circ\text{C} ΔT=0.5×102 ∘C\Delta T = 0.5 \times 10^2 \text{ }^\circ\text{C} ΔT=50 ∘C\Delta T = 50 \text{ }^\circ\text{C}

    The correct option is **B. $50^\circ

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