Answer:
Assuming that the previous example involved cooling the liquid from a certain temperature to 0°C, one way to cool the liquid 6 times as fast while starting at 0°C is to reduce the temperature of the cooling medium by a factor of 6.
For example, if the previous example involved cooling the liquid from 20°C to 0°C using a cooling medium at -10°C, and the cooling rate was not limited by the heat transfer coefficient or other factors, then one way to cool the liquid 6 times as fast while starting at 0°C is to use a cooling medium at -60°C. This would increase the temperature difference between the liquid and the cooling medium by a factor of 6, and hence increase the rate of heat transfer by the same factor.
However, it is important to note that the cooling rate of a liquid depends on many factors, such as the thermal conductivity of the liquid and the container, the surface area of the container, the volume of the liquid, and the cooling mechanism used. Therefore, other factors may need to be taken into account to achieve the desired cooling rate while maintaining the starting temperature of 0°C.