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Given density of water = 1000 kgm⁻³.

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Answer

a

2.31 litres2.31 \text{ litres}

Step-by-step solution

4 steps

Here's the solution for Item 5:

Part a· 4 steps

**Help the hotel management to determine the quantity of hot water to be given to a customer for bathing.

  1. Calculate the total mass of water required.

    The total volume of water for bathing is 10 litres10 \text{ litres}.
    Given density of water = 1000 kgm−31000 \text{ kgm}^{-3}.
    First, convert the volume to cubic meters:

    10 litres=10×10−3 m3=0.01 m310 \text{ litres} = 10 \times 10^{-3} \text{ m}^3 = 0.01 \text{ m}^3

    Now, calculate the total mass (MtotalM_{\text{total}}):

    Mtotal=Density×Volume=1000 kgm−3×0.01 m3=10 kgM_{\text{total}} = \text{Density} \times \text{Volume} = 1000 \text{ kgm}^{-3} \times 0.01 \text{ m}^3 = 10 \text{ kg}

  2. Apply the principle of calorimetry.

    Let mhm_h be the mass of hot water and mcm_c be the mass of cold water.
    The heat lost by the hot water must equal the heat gained by the cold water to reach the final temperature.

    mhc(Th−Tf)=mcc(Tf−Tc)m_h c (T_h - T_f) = m_c c (T_f - T_c)

    Where:

    • Th=72∘CT_h = 72^\circ\text{C} (hot water temperature)
    • Tc=20∘CT_c = 20^\circ\text{C} (cold water temperature)
    • Tf=32∘CT_f = 32^\circ\text{C} (final desired temperature)
    • c=4200 J kg−1 K−1c = 4200 \text{ J kg}^{-1} \text{ K}^{-1} (specific heat capacity of water, which cancels out)

  3. Set up the equation and solve for mhm_h.

    We know mc=Mtotal−mh=10 kg−mhm_c = M_{\text{total}} - m_h = 10 \text{ kg} - m_h.

    mh(72∘C−32∘C)=(10 kg−mh)(32∘C−20∘C)m_h (72^\circ\text{C} - 32^\circ\text{C}) = (10 \text{ kg} - m_h) (32^\circ\text{C} - 20^\circ\text{C}) mh(40)=(10−mh)(12)m_h (40) = (10 - m_h) (12) 40mh=120−12mh40 m_h = 120 - 12 m_h 40mh+12mh=12040 m_h + 12 m_h = 120 52mh=12052 m_h = 120 mh=12052 kg≈2.3077 kgm_h = \frac{120}{52} \text{ kg} \approx 2.3077 \text{ kg}

  4. Convert the mass of hot water to volume.

    Vh=mhDensity=2.3077 kg1000 kgm−3=0.0023077 m3V_h = \frac{m_h}{\text{Density}} = \frac{2.3077 \text{ kg}}{1000 \text{ kgm}^{-3}} = 0.0023077 \text{ m}^3

    Convert to litres:

    Vh=0.0023077 m3×1000 litres1 m3≈2.31 litresV_h = 0.0023077 \text{ m}^3 \times \frac{1000 \text{ litres}}{1 \text{ m}^3} \approx \boxed{2.31 \text{ litres}}

Part b

**Advise the hotel management on how to keep the boiled water hot for a long period of time without keeping the boiler on.

To keep boiled water hot for an extended period, the hotel management should focus on reducing heat loss. This can be achieved by:

  • Using well-insulated storage tanks (e.g., vacuum flasks or tanks with thick layers of insulating materials like foam or fiberglass) to minimize heat transfer by conduction, convection, and radiation.
  • Ensuring the tanks are tightly sealed with lids to prevent heat loss through evaporation and convection from the water surface.
  • Storing the tanks in a warm environment or away from cold drafts to reduce the temperature difference between the water and its surroundings.

Part c

**Explain to the management how the water from the boiler can reach the second floor safely.

Since carrying hot water via the staircase is not allowed, the safest and most efficient method is to use a pumping system. An electric pump can be installed to draw hot water from the ground floor boiler and push it through a network of insulated pipes up to the bathrooms on the second floor. The pipes should be well-insulated to minimize heat loss during transport. Additionally, thermostatic mixing valves should be installed at the point of use in the bathrooms. These valves automatically mix the hot water from the boiler with cold water to deliver water at a safe and comfortable bathing temperature (32∘C32^\circ\text{C}), preventing scalding.

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