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

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