Question

1. Specific heat of water is 4.184 J/g ˚C. What is the heat in KJ gained when 20 q of water is heated from 22˚C to 77˚C?

2. What is the heat in J to melt 100g of NaCl when the HF for NaCl is 518 J/g?

3. What is the heat in J to equate 50 g Acetic Acid when the HE = 390 J/g?

Answer 1

1.

Heat released / absorbed to raise the temperature of a substance from T₁ to T₂ is equal to, Q = ms(T₂ - T₁)

Where, m = mass of substance (gram)

S= specific heat (J/g ˚C)

Now,

Mass of water (m) = 20 g & s = 4.184 J/g ˚C , T₁ = 22 ˚C , T₂ = 77 ˚C

Therefore, put values in the equation

Q = 20 × 4.184 × (77 – 22 ) = 4602.4 J or 4.6024 KJ , Ans

2.

Heat of formation of NaCl (HF) = 518 J/g

Now, heat required to melt 100g of NaCl = mass × HF = 100 × 518 = 51800 J or 51.8 KJ, Ans

3.

Heat of evaporation of acetic acid = 390 J/g

Now, heat required to evaporate 100 g of acetic acid = mass × HE = 100 × 390 = 39000 J or 39 KJ, Ans