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Study the given table. a. If equal mass of these objects are heated to the same temperature and exposed in air, which one cools the fastest and why?

Physics

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Answer

b

900000 J900000 \text{ J}

Step-by-step solution

6 steps

Here are the solutions to question 20:

20. Study the given table

SubstanceSpecific heat capacity
A900 J/kg∘C900 \text{ J/kg}^\circ\text{C}
B460 J/kg∘C460 \text{ J/kg}^\circ\text{C}
C380 J/kg∘C380 \text{ J/kg}^\circ\text{C}

Part a· 3 steps

If equal mass of these objects are heated to the same temperature and exposed in air, which one cools the fastest and why?

  1. Understand specific heat capacity.

    Specific heat capacity is the amount of heat energy required to raise the temperature of 1 kg of a substance by 1∘C1^\circ\text{C}. A substance with a lower specific heat capacity requires less energy to change its temperature.

  2. Compare the specific heat capacities.

    Substance C has the lowest specific heat capacity (380 J/kg∘C380 \text{ J/kg}^\circ\text{C}) compared to Substance B (460 J/kg∘C460 \text{ J/kg}^\circ\text{C}) and Substance A (900 J/kg∘C900 \text{ J/kg}^\circ\text{C}).

  3. Determine which cools fastest.

    A substance with a lower specific heat capacity will lose heat energy more quickly for a given temperature drop when exposed to cooler surroundings. Therefore, it will cool down faster.

    • Answer: Substance C cools the fastest.
    • Reason: Substance C has the lowest specific heat capacity. This means it stores less heat energy per unit mass per degree Celsius change in temperature, and thus loses its stored heat more rapidly to the surroundings, leading to faster cooling.

Part b

900000 J900000 \text{ J}

If the temperature of substance A of 20 kg mass is increased by 50∘C50^\circ\text{C}, how much heat energy is to be supplied?

  1. Identify the given values.

    • Mass of substance A (mm) = 20 kg20 \text{ kg}
    • Change in temperature (ΔT\Delta T) = 50∘C50^\circ\text{C}
    • Specific heat capacity of substance A (cAc_A) = 900 J/kg∘C900 \text{ J/kg}^\circ\text{C} (from the table)
  2. Use the formula for heat energy.

    The amount of heat energy (QQ) supplied is given by the formula:

    Q=mcΔTQ = mc\Delta T
  3. Substitute the values and calculate.

    Q=(20 kg)×(900 J/kg∘C)×(50∘C)Q = (20 \text{ kg}) \times (900 \text{ J/kg}^\circ\text{C}) \times (50^\circ\text{C}) Q=20×900×50 JQ = 20 \times 900 \times 50 \text{ J} Q=18000×50 JQ = 18000 \times 50 \text{ J} Q=900000 JQ = 900000 \text{ J}
    • Answer: The heat energy to be supplied is 900000 J\boxed{900000 \text{ J}} (or 900 kJ900 \text{ kJ}).
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