Find the final temperature of water given an initial temperature of 28 °C, a mass of 5 kg, heat (q) of 184 J. (Specific Heat of water = 4.184). a. 19.2°C b. 36.8 °C c. 38.8 °C d. 3880 °C

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

answerto predict;