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

sup bro, im dumb, but this is what i found xd

"the ozone layer acts as a filter for the shorter wavelength and highly hazardous ultraviolet radiation (uvr) from the sun, protecting life on earth from its potentially harmful effects. when the sky is clear, there is an inverse relationship between stratospheric ozone and solar uvr measured at the earth's surface"