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A 100g sample of steam at 100°C is passed into 500g of water at 20°C. Calculate the final temperature of the mixture. (Specific heat capacity of water = 4200 J/kg K, Latent heat of vaporization of ste

Physics

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Answer

C

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26.

  1. State the principle of calorimetry.

    Heat lost by steam = Heat gained by water.
    We need to consider two phases for the steam: condensation and cooling of the condensed water.
    Let TfT_f be the final temperature.

  2. Calculate the maximum heat that can be gained by the water to reach 100∘C100^\circ\text{C}.

    The initial temperature of water is 20∘C20^\circ\text{C}. If the final temperature is 100∘C100^\circ\text{C}, the change in temperature for water is ΔTw=100∘C−20∘C=80∘C\Delta T_w = 100^\circ\text{C} - 20^\circ\text{C} = 80^\circ\text{C}.
    Given:

    • Mass of water (mwm_w) = 500 g=0.5 kg500 \text{ g} = 0.5 \text{ kg}
    • Specific heat capacity of water (cwc_w) = 4200 J/kg K4200 \text{ J/kg K}

    Heat gained by water (Qgained,waterQ_{gained, water}) to reach 100∘C100^\circ\text{C}:

    Qgained,water=mwcwΔTwQ_{gained, water} = m_w c_w \Delta T_w Qgained,water=(0.5 kg)×(4200 J/kg K)×(80 K)Q_{gained, water} = (0.5 \text{ kg}) \times (4200 \text{ J/kg K}) \times (80 \text{ K}) Qgained,water=168000 JQ_{gained, water} = 168000 \text{ J}
  3. Calculate the maximum heat that can be lost by the steam if all of it condenses and cools to 20∘C20^\circ\text{C}.

    This involves two parts:

    • Heat lost by condensation of steam at 100∘C100^\circ\text{C} to water at 100∘C100^\circ\text{C} (QcondenseQ_{condense}).
    • Heat lost by the condensed water (originally steam) from 100∘C100^\circ\text{C} to 20∘C20^\circ\text{C} (Qcool,steamQ_{cool, steam}).

    Given:

    • Mass of steam (msm_s) = 100 g=0.1 kg100 \text{ g} = 0.1 \text{ kg}
    • Latent heat of vaporization (LvL_v) = 2.26×106 J/kg2.26 \times 10^6 \text{ J/kg}

    Heat lost by condensation:

    Qcondense=msLvQ_{condense} = m_s L_v Qcondense=(0.1 kg)×(2.26×106 J/kg)Q_{condense} = (0.1 \text{ kg}) \times (2.26 \times 10^6 \text{ J/kg}) Qcondense=226000 JQ_{condense} = 226000 \text{ J}
  4. Compare the heat values to determine the final temperature.

    The heat released by just the condensation of all the steam (226000 J226000 \text{ J}) is greater than the heat required to raise the temperature of the water to 100∘C100^\circ\text{C} (168000 J168000 \text{ J}).
    This means that the water will reach 100∘C100^\circ\text{C}, and not all the steam will condense. The excess heat from the steam (if it were to cool further) is not absorbed by the water because the water itself reaches the boiling point of the steam. Therefore, the final temperature of the mixture will be 100∘C100^\circ\text{C}.

    To confirm, the mass of steam required to condense to bring the water to 100∘C100^\circ\text{C} is:

    mcondensed=Qgained,waterLv=168000 J2.26×106 J/kg≈0.0743 kg=74.3 gm_{condensed} = \frac{Q_{gained, water}}{L_v} = \frac{168000 \text{ J}}{2.26 \times 10^6 \text{ J/kg}} \approx 0.0743 \text{ kg} = 74.3 \text{ g}

    Since 74.3 g74.3 \text{ g} is less than the available 100 g100 \text{ g} of steam, the water will reach 100∘C100^\circ\text{C}, and some steam will remain uncondensed at 100∘C100^\circ\text{C}.

    The correct option is C.

    C. 100∘C\boxed{\text{C. } 100^\circ\text{C}}

    27.
    Good insulators are materials that trap air, which is a poor conductor. Air is an excellent thermal insulator because its molecules are far apart, reducing the frequency of collisions needed for heat transfer by conduction. Many insulating materials, like fiberglass, foam, or wool, work by trapping pockets of air.

    The correct option is B.

    B. Trap air, which is a poor conductor\boxed{\text{B. Trap air, which is a poor conductor}}
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