Find the increase in entropy of 1.00 kg of ice originally at 0 degrees Centigrade, that is melted to form water

Physics

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Answer

ΔS=1223 J/K\Delta S = 1223 \text{ J/K}

Step-by-step solution

3 steps

Here's how to calculate the increase in entropy:

To find the increase in entropy (ΔS\Delta S) during a phase change at constant temperature, we use the formula:

ΔS=QT\Delta S = \frac{Q}{T}

where QQ is the heat absorbed and TT is the absolute temperature in Kelvin.

For melting, the heat absorbed (QQ) is given by:

Q=mLfQ = mL_f

where mm is the mass and LfL_f is the latent heat of fusion.

Given values:

  • Mass of ice, m=1.00 kgm = 1.00 \text{ kg}
  • Temperature, TC=0∘CT_C = 0^\circ \text{C}
  • Latent heat of fusion of water, Lf=334×103 J/kgL_f = 334 \times 10^3 \text{ J/kg} (standard value)
  1. Convert the temperature from Celsius to Kelvin.

    TK=TC+273.15T_K = T_C + 273.15 TK=0+273.15=273.15 KT_K = 0 + 273.15 = 273.15 \text{ K}
  2. Calculate the heat absorbed (QQ) to melt the ice.

    Q=mLfQ = mL_f Q=(1.00 kg)×(334×103 J/kg)Q = (1.00 \text{ kg}) \times (334 \times 10^3 \text{ J/kg}) Q=334×103 JQ = 334 \times 10^3 \text{ J}
  3. Calculate the increase in entropy (ΔS\Delta S).

    ΔS=QTK\Delta S = \frac{Q}{T_K} ΔS=334×103 J273.15 K\Delta S = \frac{334 \times 10^3 \text{ J}}{273.15 \text{ K}} ΔS≈1222.84 J/K\Delta S \approx 1222.84 \text{ J/K} ΔS=1223 J/K\Delta S = 1223 \text{ J/K}
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