When liquid oxygen at $50 \text{ K}$ is heated, its temperature will initially rise until it reaches its boiling point (which is approximately $90 \text{ K}$). During this period, the specific heat of the liquid determines the slope of the temperature-time graph. Once it reaches the boiling point, a phase change from liquid to gas occurs. During this phase transition, the temperature remains constant despite continuous heating, as the heat supplied is used as the latent heat of vaporization. This corresponds to a horizontal line (plateau) on the graph. After all the liquid oxygen has vaporised into gas, the continuous heating will cause the temperature of the gaseous oxygen to rise steadily up to $300 \text{ K}$. Therefore, the correct graph must show an initial positive slope, a horizontal plateau at the boiling point, and then another positive slope.