Question

100.-gram sample of H2O() at 22.0°C absorbs 8360 joules of heat. What will be the final temperature of the water?

Answer 1

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Assume Specific Heat capacity of water = 4.186 J/g°C
100g sample, 
Q=mcΔθ
8360=100(4.186)(Final Temp.-22.0)
Final Temperature ≈41.97°C

Answer 2

Answer:

Final Temperature =  41.9°C

Explanation:

Mass = 100g

Initial Temperature (T1) = 22°C

Final Temperature (T2) = ?

Heat = 8360

Formular relating these parameters is given as;

H = MC(T2-T1)

Where C = Specific Heat Capacity of water = 4.2 J/g°C

Inserting the values, we have;

8360 = 100 * 4.2 * (T2 - 22)

T2 - T1 = 8360 / 420

T2 - T1 = 19.9

T2 = T1 + 19.9

T2 = 22 + 19.9

T2 = 41.9°C