Calculate the amount of heat released when the bottle (250g) of water is cooled from 30°C to 25°C. The specific heat of water is 4.186J/g Degree Celsius

5232.5J

Explanation:

Data obtained from the question. This includes the following:

Mass (M) = 250g

Initial temperature (T1) = 30°C

Final temperature (T2) = 25°C

Change in temperature (ΔT) = T1 - T2 = 30°C - 25°C = 5°C

Specific heat capacity (C) = 4.186J/g°C

Heat (Q) =.?

The heat released can be obtained as follow:

Q = MCΔT

Q = 250 x 4.186 x 5

Q = 5232.5J

Therefore, the heat released is 5232.5J