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A mass of 75 g of steam at 100°C condenses into water at 100°C. Calculate the mass of ice at 0°C that this steam can just melt.

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2 parts
a

507 g\text{507 g}

b

12.9∘C\text{12.9}^\circ \text{C}

Step-by-step solution

6 steps

Here's how to solve the problem:

Part a· 3 steps

To calculate the mass of the ice, we use the principle of conservation of energy. The heat released by the steam condensing is equal to the heat absorbed by the ice melting.

Given:
Mass of steam (msm_s) = 75 g=0.075 kg75 \text{ g} = 0.075 \text{ kg}
Initial temperature of steam = 100∘C100^\circ \text{C}
Initial temperature of melting ice = 0∘C0^\circ \text{C}
Latent heat of vaporization of water (LvL_v) = 2.26×106 J/kg2.26 \times 10^6 \text{ J/kg}
Latent heat of fusion of ice (LfL_f) = 3.34×105 J/kg3.34 \times 10^5 \text{ J/kg}

  1. Calculate the heat released by the steam as it condenses to water at 100∘C100^\circ \text{C}.

    Qsteam=msLvQ_{\text{steam}} = m_s L_v Qsteam=(0.075 kg)×(2.26×106 J/kg)Q_{\text{steam}} = (0.075 \text{ kg}) \times (2.26 \times 10^6 \text{ J/kg}) Qsteam=169500 JQ_{\text{steam}} = 169500 \text{ J}

  2. Calculate the heat required to melt the ice at 0∘C0^\circ \text{C}.

    Let mim_i be the mass of the ice.

    Qice=miLfQ_{\text{ice}} = m_i L_f Qice=mi×(3.34×105 J/kg)Q_{\text{ice}} = m_i \times (3.34 \times 10^5 \text{ J/kg})

  3. Equate the heat released by the steam to the heat absorbed by the ice, as the steam's heat is "just sufficient to melt the ice."

    Qsteam=QiceQ_{\text{steam}} = Q_{\text{ice}} 169500 J=mi×(3.34×105 J/kg)169500 \text{ J} = m_i \times (3.34 \times 10^5 \text{ J/kg}) mi=169500 J3.34×105 J/kgm_i = \frac{169500 \text{ J}}{3.34 \times 10^5 \text{ J/kg}} mi≈0.507485 kgm_i \approx 0.507485 \text{ kg} mi≈507.485 gm_i \approx 507.485 \text{ g}

    The mass of the ice is 507 g\boxed{\text{507 g}}.

Part b

12.9∘C\text{12.9}^\circ \text{C}

To calculate the equilibrium temperature of the mixture, we consider the two masses of water that result from the phase changes

  1. 0.075 kg0.075 \text{ kg} of water at 100∘C100^\circ \text{C} (from the condensed steam).
  2. 0.507485 kg0.507485 \text{ kg} of water at 0∘C0^\circ \text{C} (from the melted ice).

We will use the specific heat capacity of water (cw=4186 J kg−1 K−1c_w = 4186 \text{ J kg}^{-1} \text{ K}^{-1}).

  1. Apply the principle of calorimetry

    Heat lost by hot water = Heat gained by cold water.
    Let TfT_f be the final equilibrium temperature.
    Heat lost by water from condensed steam: Qlost=mscw(100∘C−Tf)Q_{\text{lost}} = m_s c_w (100^\circ \text{C} - T_f)
    Heat gained by water from melted ice: Qgained=micw(Tf−0∘C)Q_{\text{gained}} = m_i c_w (T_f - 0^\circ \text{C})

  2. Equate the heat lost and heat gained.

    The specific heat capacity cwc_w cancels out.

    mscw(100−Tf)=micw(Tf−0)m_s c_w (100 - T_f) = m_i c_w (T_f - 0) ms(100−Tf)=miTfm_s (100 - T_f) = m_i T_f

  3. Substitute the known values and solve for TfT_f.

    (0.075 kg)(100−Tf)=(0.507485 kg)Tf(0.075 \text{ kg}) (100 - T_f) = (0.507485 \text{ kg}) T_f 7.5−0.075Tf=0.507485Tf7.5 - 0.075 T_f = 0.507485 T_f 7.5=0.507485Tf+0.075Tf7.5 = 0.507485 T_f + 0.075 T_f 7.5=(0.507485+0.075)Tf7.5 = (0.507485 + 0.075) T_f 7.5=0.582485Tf7.5 = 0.582485 T_f Tf=7.50.582485T_f = \frac{7.5}{0.582485} Tf≈12.876∘CT_f \approx 12.876^\circ \text{C}

    The equilibrium temperature of the mixture is 12.9∘C\boxed{\text{12.9}^\circ \text{C}}.

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