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
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Answer:
Try this link https://chem.libretexts.org/Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/08%3A_Solids_Liquids_and_Gases/8.E%3A_Solids_Liquids_and_Gases_(Exercises)
Latent heat of fusion for ice is 33600J/k
Heat absorbed
- k_f×m
- 31.6×33600
- 1061,740mJ
- 1061J
Answer:
2584 g
Explanation:
The molecular formula for sucrose is C₁₂H₂₂O₁₁, which means its molar mass is 342.3 g/mol.
First we<u> convert molecules into moles</u> by using <em>Avogadro's number</em>:
- 4.55x10²⁴ molecules ÷ 6.023x10²³ mol/molecules = 7.55 mol
Now we <u>convert moles of sucrose into grams</u>, using its<em> molar mass</em>:
- 7.55 mol * 342.3 g/mol = 2584 g
