To determine the temperature at which the equilibrium between solid water (ice) and liquid water exists at a pressure of 101.3 kilopascals (standard atmospheric pressure), we need to recall that this scenario represents the melting or freezing point of water.
Water (H₂O) exists in solid form (ice) and liquid form and can transition between these states. The specific temperature at which pure water reaches an equilibrium between its solid phase (ice) and liquid phase is a well-known physical fact.
1. Option 1: [tex]\(0 K\)[/tex]
Zero Kelvin (0 K) is the absolute zero temperature, the point where molecular motion theoretically stops. This temperature is far too low for water to exist in liquid form. At 0 K, all substances would be solid.
2. Option 2: [tex]\(0^{\circ} C\)[/tex]
Zero degrees Celsius (0°C) is the temperature at which water freezes or melts at 101.3 kPa. At this temperature, ice and liquid water exist in equilibrium.
3. Option 3: [tex]\(32 K\)[/tex]
Thirty-two Kelvin (32 K) is a very low temperature, much lower than water's freezing point. At 32 K, water would be far below freezing and exist only as ice.
4. Option 4: [tex]\(273^{\circ} C\)[/tex]
Two hundred seventy-three degrees Celsius (273°C) is significantly higher than the boiling point of water at standard atmospheric pressure (100°C). At this temperature, water would be in vapor form, not in equilibrium between solid and liquid states.
Considering this information, the temperature at which the equilibrium:
[tex]\[ H _2 O ( s ) \rightleftharpoons H _2 O (\ell) \][/tex]
exists at 101.3 kilopascals is [tex]\(0^{\circ} C\)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{0^{\circ} C}
\][/tex]
