Answer:
238,485 Joules
Explanation:
The amount of energy required is a summation of heat of fusion, capacity and vaporization.
Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)
m (mass of water) = 75 g
Lf (specific latent heat of fusion of water) = 336 J/g
C (specific heat capacity of water) = 4.2 J/g°C
∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C
Lv (specific latent heat of vaporization of water) = 2,260 J/g
Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J
Answer:
I know that the 100-mL graduated cylinders are always read to 1 decimal place.
I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.
