A 25 g sample of water releases 2913 J as it cools from 50.0°C. Calculate the final temperature of the water. (Water’s specific heat capacity is 4.184 J/g°C) (round answer to 2-3 sig figs)

The final temperature is 22.2°C

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass (M) = 25g

Heat (Q) = 2913 J

Initial temperature (T1) = 50.0°C

Final temperature (T2) =?

Specific heat capacity (C) = 4.184J/g°C

Step 2:

Determination of the final temperature.

The final temperature can be obtained as follow:

Q = MC(T1 – T2).. Since we are cooling

2913 = 25 x 4.184 (50 – T2)

2913 = 104.6 (50 – T2)

Divide both side by 104.6

50 – T2 = 2913/104.6

50 – T2 = 27.8

Collect like terms

T2 = 50 – 27.8

T2 = 22.2°C

The final temperature is 22.2°C

fire could start, buildings collapsing, and ash falling from the sky.

explanation:

this is because the buildings or plants could catch on fire and the buildings would most likely collapse. lastly ash would fall because all the burning would cause ash to fall from the sky.

hope this .