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Find the final temperature of water given:

- Initial temperature: 28 °C
- Mass: 5 kg
- Heat (\(q\)): 184 J
- Specific Heat of water: 4.184 J/g°C

A. 19.2 °C
B. 36.8 °C
C. 38.8 °C
D. 3880 °C

Answer :

Explanation:

Initial temp = 28 °C

Final Temp = ?

Heat = 184 J

mass = 5 g

c = 4.184 joule/gram °C

The relationship between these parameters is given by the equation;

H = mCΔT

ΔT = H / mC

ΔT = 184 / 5(4.184)

ΔT = 184 / 20.92

ΔT = 8.8 °C

ΔT = Final Temp - Initial Temp

Final Temp = Initial Temp + ΔT

Final Temp = 28°C + 8.8 °C

Final Temp = 36.8°C

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