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Discuss two factors for choosing cooking fuel and calculate the heat evolved in an experiment with given water volume and temperature change.

Physics

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Answer

b(i)

39.825 kJ\text{39.825 kJ}

Step-by-step solution

3 steps

Part a

Two factors that should be considered when choosing fuel for cooking are

  • Calorific value: The amount of heat produced per unit mass or volume of the fuel.
  • Cost and availability: The economic feasibility and ease of obtaining the fuel.

Part b(i)

39.825 kJ\text{39.825 kJ}

Heat evolved during the experiment.

Given data:
Volume of water (VV) = 450 cm3450 \text{ cm}^3
Initial temperature of water (TiT_i) = 25∘C25^\circ \text{C}
Final temperature of water (TfT_f) = 46.5∘C46.5^\circ \text{C}
Density of water (ρ\rho) = 1 g/cm31 \text{ g/cm}^3
Specific heat capacity of water (cc) = 4.2 kJ/kg K4.2 \text{ kJ/kg K} (assuming the given 4.2 kg/kg K4.2 \text{ kg/kg K} is a typo and should be 4.2 kJ/kg K4.2 \text{ kJ/kg K})

  1. Calculate the mass of water.

    The mass of water can be calculated using its density and volume.

    m=ρ×Vm = \rho \times V m=1 g/cm3×450 cm3m = 1 \text{ g/cm}^3 \times 450 \text{ cm}^3 m=450 gm = 450 \text{ g}

    Convert mass to kilograms:

    m=450 g×1 kg1000 gm = 450 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} m=0.450 kgm = 0.450 \text{ kg}
  2. Calculate the change in temperature of the water.

    ΔT=Tf−Ti\Delta T = T_f - T_i ΔT=46.5∘C−25∘C\Delta T = 46.5^\circ \text{C} - 25^\circ \text{C} ΔT=21.5∘C\Delta T = 21.5^\circ \text{C}

    Note that a change in temperature of 21.5∘C21.5^\circ \text{C} is equivalent to a change of 21.5 K21.5 \text{ K}.

  3. Calculate the heat evolved (heat absorbed by the water).

    The heat absorbed by the water (QQ) is calculated using the formula Q=mcΔTQ = mc\Delta T.

    Q=m×c×ΔTQ = m \times c \times \Delta T Q=0.450 kg×4.2 kJ/kg K×21.5 KQ = 0.450 \text{ kg} \times 4.2 \text{ kJ/kg K} \times 21.5 \text{ K} Q=39.825 kJQ = 39.825 \text{ kJ}

    The heat evolved during the experiment is 39.825 kJ\boxed{\text{39.825 kJ}}.

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