How much heat is required to heat 1.6g of ice from -16c to steam at 112c?

Total heat ≅ 49.07 kJ

Explanation:

Given that:

mass = 1.6 g = 0.016 kg

Initial temperature = - 16 ° C

final temperature = 112° C

specific heat for ice = 2.06 kJ/kgC

specific heat of water = 4.186 kJ/kgC

heat fusion of ice = 334 kJ/kg

specific heat for steam = 2.1 kJ/kgK

heat of vaporization of water = 2256 kJ/kg

To heat ice from -16 ° C to 0 ° C

Q₁ =  2.06 kJ/kgC × 0.016 kg ×  16 ° C

Q₁ =  0.52736 kJ

To melt Ice at 0° C

Q₂= 334 kJ/kg × 0.016 kg = 5.344 kJ

To heat water from 0° C to  100° C

Q₃ = 4.186 kJ/kgC × 0.016 kg  × 100° C

Q₃ = 6.6976 kJ

To vaporize water to steam at 100° C

Q₄ = 2256 kJ/kg × 0.016 kg = 36.096 kJ

To heat steam from 100C to 112° C

Q₅ = 2.1 kJ/kgC × 0.016 kg × 12 C

Q₅ = 0.4032 kJ

Total heat = Q₁ + Q₂ + Q₃ + Q₄  + Q₅

Total heat =  (0.52736 +  5.344 +  6.6976 + 36.096 + 0.4032) kJ

Total heat = 49.06816  kJ

Total heat ≅ 49.07 kJ

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

i think it's letter a.