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A piece of metal of mass 50g is cooled from 80°C to 20°C. Calculate the amount of heat lost. (Specific heat capacity of the material of metal = 450 Jkg-1k-1. A. 4.50 x 103J. B. 2.5 x 103J. C. 1.80 x 103J. D. 1.35 x 103J. [2000/23])

Physics

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Step-by-step solution

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To calculate the amount of heat lost, we use the formula Q=mcΔTQ = mc\Delta T.

  1. Identify the given values and convert units.

    Mass of metal, m=50 g=0.050 kgm = 50 \text{ g} = 0.050 \text{ kg}
    Initial temperature, T1=80 °CT_1 = 80 \text{ °C}
    Final temperature, T2=20 °CT_2 = 20 \text{ °C}
    Specific heat capacity, c=450 J kg−1 K−1c = 450 \text{ J kg}^{-1} \text{ K}^{-1}

  2. Calculate the change in temperature (ΔT\Delta T).

    The change in temperature is the difference between the initial and final temperatures.

    ΔT=T1−T2=80 °C−20 °C=60 °C\Delta T = T_1 - T_2 = 80 \text{ °C} - 20 \text{ °C} = 60 \text{ °C}

    Since a change of 1 °C is equal to a change of 1 K, ΔT=60 K\Delta T = 60 \text{ K}.

  3. Substitute the values into the heat transfer formula.

    Q=mcΔTQ = mc\Delta T Q=(0.050 kg)×(450 J kg−1 K−1)×(60 K)Q = (0.050 \text{ kg}) \times (450 \text{ J kg}^{-1} \text{ K}^{-1}) \times (60 \text{ K})

  4. Calculate the amount of heat lost.

    Q=0.050×450×60 JQ = 0.050 \times 450 \times 60 \text{ J} Q=22.5×60 JQ = 22.5 \times 60 \text{ J} Q=1350 JQ = 1350 \text{ J}

  5. Express the answer in scientific notation and compare with the options.

    Q=1.35×103 JQ = 1.35 \times 10^3 \text{ J}

    This matches option D.

    The final answer is D. 1.35×103 J\boxed{\text{D. } 1.35 \times 10^3 \text{ J}}.

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