Ice Built Up in a Freezer

Task number: 1281

Assume that every time you open the freezer with a volume of 210 l, it fills with air with a temperature of 25 °C and a relative humidity of 80 %. How much ice builds up in it after it is once opened, given that the air cools down to a temperature of -20 °C? Compare with the mass of water that is created in a refrigerator with inner temperature of 5 °C. Try to estimate the amount of the heat released in both cases.

  • Hint – the absolute and relative humidity

    At a certain temperature the air “holds” (without condensation) a certain amount of water. This maximum amount is referred to as the saturated water vapor and we can find its density for a given temperature in The Handbook of Chemistry and Physics. Typically, however, the air contains less vapor. The amount can be determined by the density of water vapor at a specific moment (the absolute humidity of air) or by stating what part of the water vapor (usually percentage) of the maximum possible amount for a given temperature is currently contained in the air (the so-called relative humidity of air). The density of saturated water vapor decreases with the decreasing temperature, in other words at a lower temperature the air “holds less water”.

  • Hint – how to solve this problem

    When you open the freezer, the warm and humid air from the room gets inside. The air cools down and part of the water vapor condenses and consequently freezes.

    To determine the amount of water vapor in the air at a given temperature, use the density of the saturated water vapor, which can be found in The Handbook of Chemistry and Physics.

  • Numerical values

    V = 210 l = 0.210 m3  the internal volume of the freezer
    t1 = 25 °Cthe room air temperature
    φ = 80 %the relative humidity of air in the room
    t2 = −20 °Cthe temperature in the freezer
    t3 = 5  °Cthe temperature in the refrigerator
    mF = ?the mass of built-up ice in the freezer
    mR = ?the mass of built-up ice in the refrigerator
    QF = ?the heat released in the freezer
    QR = ?the heat released in the refrigerator

    Other necessary values:

    ρ1 = 23 g/m3 = 2,3×10−2 kg/m3the density of the saturated water vapor at the temperature of 25 °C
    ρ2 = 0.88 g/m3 = 8.8×10−4 kg/m3the density of the saturated water vapor at the temperature of −20 °C
    ρ3 = 6.8 g/m3 = 6.8×10−3 kg/m3the density of the saturated water vapor at the temperature of 5 °C
    tm = 0 °Cthe melting point of ice
    cI = 2.1 kJkg−1K−1 = 2100 Jkg−1K−1the specific heat capacity of ice
    lfus = 334 kJkg−1 = 3.34×105 Jkg−1the specific latent heat of fusion of ice
    cW = 4.18 kJkg−1K−1 = 4180 Jkg−1K−1  the specific heat capacity of water
    lvap = 2260 kJkg−1 = 2.26×106 Jkg−1the specific latent heat of vaporization of water
  • Analysis

    At a certain temperature the air “holds” (without condensation) a certain amount of water. This maximum amount is referred to as the saturated water vapor and we can find its density for a given temperature in The Handbook of Chemistry and Physics. Typically, however, the air contains less water. The amount can be determined by the density of water vapor at a specific moment (the absolute humidity of air) or by stating what part of the water vapor (usually percentage) of the maximum possible amount for a given temperature is currently contained in the air (the so-called relative humidity of air). With decreasing temperature decreases also the density of saturated water vapor, that is, at a lower temperature the air “holds less water” (see above).

    We look up the density of the saturated water vapor for the room temperature in The Handbook of Chemistry and Physics, and according to the given relative humidity of air we determine the density of the water vapor in the room. Similarly, we look up the density of the saturated water vapor for the temperature in the freezer. We determine the mass of the water vapor, which is shut in the freezer, and the mass of the water vapor that can be contained in the cold air in the freezer. The remaining water vapor first condenses, then cools down, freezes and consequently cools down even more.

  • Solution

    First, we determine the density ρr of the water vapor in the room. The following relation applies according to the definition of the relative humidity:

    \[\varphi _1 = \frac{ \varrho _r}{\varrho _1} \Rightarrow \] \[\Rightarrow \varrho _r = \varphi_1 \varrho_1 = 0.8 \cdot{2.3} \cdot{10^{-2}} \,\mathrm kgm^{-3}=\] \[\,\ \,\ = 1.84 \cdot{10^{-2}} \,\mathrm kgm^{-3}\]

    This means that when we open the freezer, the water vapor with a mass of \(m_1 = \varrho_r V\) gets inside along with the air.

    Having a saturated water vapor in the freezer means that the inside air can “hold” only a mass \(m_2 = \varrho_2 V\) of the water vapor. The excess water vapor condenses and freezes, i.e. the amount of ice that is created when we open the freezer equals the difference between these two masses:

    \[m_F = m_1 - m_2 = \varrho_r V - \varrho_2 V = (\varphi_1 \varrho_1 - \varrho_2)V\]

    \[m_F = (0.8 \cdot{2.3}\cdot{10^{-2}} - 8.8 \cdot{10^{-4}})\cdot 0.21 \,\mathrm kg \dot= 3.7\cdot{10^{-3}} \,\mathrm kg =\] \[\,\ \ \,= 3.7 \,\mathrm g\]

    For the refrigerator, it is necessary to use the different density of the saturated water vapor \(\varrho_3\):

    \[m_{R}=(\varphi_1 \varrho_1 - \varrho_3)V\]

    \[m_{R}= (0.8 \cdot{2.3}\cdot{10^{-2}} - 6.8 \cdot{10^{-3}})\cdot 0.21 \,\mathrm kg \dot=\] \[\,\ \ \ \dot= 2.4\cdot{10^{-3}} \,\mathrm kg = 2.4 \,\mathrm g\]

    When estimating the released heat we consider that in the freezer the water vapor first condenses, then the water cools down, freezes and the ice cools down to the inner temperature of the freezer. As we do not know the specific latent heat of condensation, we use the specific latent heat of vaporization. The amount of the released heat will be, in fact, greater.

    \[Q_F = l_{vap} m_F + c_W m_{F} (t_1 - t_m) + l_{fus} m_F + c_Im_F (t_m-t_2)=\] \[=\ \,\ \, [l_{vap} + c_W(t_1-t_m) + l_{fus} + c_I(t_m-t_2)]m_I\]

    \[Q_F = [ 2.26 \cdot{10^6} + 4180 \cdot{(25-0)} + 3.34\cdot{10^5} + 2100\cdot (0+20)] \cdot 3.7 \cdot{10^{-3}}\dot=\] \[\, \ \ \,\, \ \dot= 10000 \,\mathrm J= 10 \, \mathrm kJ \]

    The water vapor in the refrigerator first condenses and then the water cools down:

    \[Q_{R} = l_{vap} m_{R} + c_W m_{R} (t_1 - t_3) = [l_{vap} + c_W(t_1-t_3)]m_{R}\]

    \[Q_{R} = [2.26 \cdot{10^6} + 4180 \cdot (25-5)]\cdot 2.4 \cdot{10^{-3}} \dot= 5600\,\mathrm J = 5.6 \,\mathrm kJ\]

  • Answer

    When we open the freezer, the ice with a mass of 3.7 g is built up. This releases at least 10 kJ of heat. In the case of the refrigerator the water with a mass of 2.4 g is created, which releases the amount of 5.6 kJ of heat.

Difficulty level: Level 2 – Upper secondary level
Send comment on task by email