The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
to learn more about energy
brainly.com/question/7185299
#SPJ4
I think its either A or D
Answer:
<h2>The answer is 14.29 %</h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula
From the question
actual density = 0.70 g/mL
error = 0.8 - 0.7 = 0.1
So we have
We have the final answer as
<h3>14.29 %</h3>
Hope this helps you
