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The question asks how much energy is transferred from the thermal energy store of the water as it cools.

The initial statement tells us that it takes 90 J to increase the temperature of water by 6°C. Therefore, if the water were cooled by 6°C, 90 J of energy would be transferred from the water. However, the water only cools by 3°C, which is half of the original 6°C change. 

Since the temperature change is halved, the energy transferred will also be halved. So, half of 90 J is:

\[
\frac{90}{2} = 45 \, J
\]

Thus, 45 J of energy is transferred as the water cools by 3°C.

Would you like further details on this calculation or any related concepts? Here are 5 related questions you might find helpful:

1. How is energy transfer related to temperature change?
2. What is the specific heat capacity and how does it apply to this scenario?
3. How would the energy transfer change if the water cooled by a different amount?
4. What formula can be used to calculate energy transfer in thermal processes?
5. How does the energy transfer in water compare to other materials?

**Tip:** When working with energy transfer in temperature changes, always check the proportion of temperature change to the energy needed, as it can scale linearly if no phase change is involved.

By how many joules does the energy in the thermal energy store of the water change as it cools?

Learn how to calculate the energy transferred when water cools by 3°C. Based on the fact that 90 J is required to heat the water by 6°C, we determine the energy transfer proportionally. Suitable for middle school students (Grades 6-8).

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