I believe the answer would be 446.9 J.
The latent heat of fusion of water is 80 cal/g.
The specific heat of water is 1 cal/g-C.
The latent heat of vaporization of water is 540 cal/g.
Therefore, if we have 1.5 kg = 1500 g, the total heat requirement is:
1500 g[(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g(720 cal/g) = 1,080,000 cal.
Answer:
Distance in mm will be 0.3718 mm
Explanation:
We have given charge surface charge density 
We know that electric field due to surface charge density is given by
We have given potential difference V = 105 volt
We know that potential difference is given by 
So 
<span>Nuclear fission produces elements that are heavier than helium.
The elements that are used in nuclear fission and their products are much heavier than helium.</span>
B. the sum of the protons and the neutrons in one atom of the element
