Final answer:
When 25 grams of aluminum cools from 55°C to 25°C, 675 joules of heat is given off. This is calculated using the formula for heat transfer, which in this case results in -675 J, representing the heat released, thus the positive absolute value is taken.
Step-by-step explanation:
The question is asking how much heat is released when a mass of aluminum cools down from a higher temperature to a lower temperature. The formula to calculate the heat ($Q$), given off by a substance when it changes temperature, is $Q = m \times c \times \Delta T$, where:
- $m$ is the mass of the substance,
- $c$ is the specific heat of the substance, and
- $\Delta T$ is the change in temperature.
In this case, for aluminum with a specific heat of 0.90 J/g°C, a mass of 25 grams, a starting temperature of 55°C, and an ending temperature of 25°C, the heat given off can be calculated as follows:
$Q = 25 \text{ g} \times 0.90 \text{ J/g°C} \times (25°C - 55°C) = 25 \times 0.90 \times (-30) = -675 \text{ J}$
Since the heat energy released is a positive value, the absolute value of $-675$ J is $675$ J. Therefore, the correct answer is that 675 joules of heat is given off when 25 grams of aluminum is cooled from 55°C to 25°C.