Which provides more cooling for a styrofoam cooler, one with 10 lbs of ice at 0 degrees C or one with 10 lbs of ice water at 0 degrees C. Explain.

qice=mass*heat of fusion.
qH2O=mass*specific heat x delta T

However, you aren't given a two temperatures; therefore, you can't actually calculate q. You must reason out the answer.
10 lbs ice will give the cooling power for the heat of fusion to melt the ice PLUS you will have 10 lbs water at 0 degrees C to cool.
For the 10 lbs water at 0 degrees C you have just the water to act as a coolant. Given this information, pick which you think will have the greatest cooling power.

The 10 lbs of ice at 0 degrees C will provide more cooling for the styrofoam cooler because it will provide the cooling power for the heat of fusion to melt the ice PLUS you will have 10 lbs of water at 0 degrees C to cool.

To determine which provides more cooling for a styrofoam cooler, we need to compare the two scenarios: one with 10 lbs of ice at 0 degrees Celsius and the other with 10 lbs of ice water at 0 degrees Celsius.

In the first scenario, you have 10 lbs of ice at 0 degrees Celsius. The cooling power of the ice comes from the heat of fusion, which is the amount of heat energy required to change a substance from a solid to a liquid without changing its temperature. The equation you mentioned, qice = mass * heat of fusion, can be used to calculate this cooling power. However, since you don't have information about the change in temperature, you can't calculate the exact value of q. But keep in mind that the ice will provide cooling not only during the melting process, but also in its solid state.

In the second scenario, you have 10 lbs of ice water at 0 degrees Celsius. Here, the cooling power comes from the specific heat of water, which is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius. The equation you mentioned, qH2O = mass * specific heat * delta T, can be used to calculate the cooling power. However, since the temperature is already at 0 degrees Celsius, there won't be any temperature change (delta T = 0). Therefore, the cooling power in this case relies solely on the water itself as a coolant.

Comparing the two scenarios, the first one with 10 lbs of ice provides not only the cooling power from the heat of fusion, but also the benefit of having the melted ice water as an additional coolant. On the other hand, the second scenario with 10 lbs of ice water only relies on the water as a coolant since there is no further temperature change.

Based on this reasoning, it is likely that the scenario with 10 lbs of ice will provide more cooling for the styrofoam cooler compared to the scenario with 10 lbs of ice water at 0 degrees Celsius.

The styrofoam cooler with 10 lbs of ice at 0 degrees C will provide more cooling than the one with 10 lbs of ice water at 0 degrees C.

This is because the 10 lbs of ice will have both the cooling power from the heat of fusion to melt the ice, as well as the 10 lbs of water at 0 degrees C to provide additional cooling. On the other hand, the 10 lbs of ice water only has the water itself to act as a coolant.

Overall, the presence of the solid ice in the first cooler gives it more cooling power compared to the second cooler with ice water.