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
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Hope this helps you :)
Answer:
Using the coarse adjustment knob of the microscope in high power may lead to the breaking of the slide if adjusted and raised the slide too much which can damage the sample as well as the high power lens.
In this case, I would recommend using the fine adjustment knob and moving away from the end of the viewing area of the microscope so there would no collision take place. The fine adjustment will help to get a clear image.
