Calculate the heat energy required to vaporize 50g of water initially at 80⁰C if the specific heat capacity of water is 4.23jg⁻¹k⁻¹ (latent heat of vaporization of water is 2260jg⁻¹)
Heat needed to vaporize water = heat acquired by water + heat acquired by steam
H = MCΔT + ML
H = M(CΔT + L)
M = mass of water = 50g
C = specific heat capacity of water = 4.23jg⁻¹k⁻¹
ΔT = change in temperature = 100 – 80 = 20⁰C = 273 + 20 = 293
L = 2260jg⁻¹
Substituting all these, we have:
H = 50((4.23 × 293) + 2260)
H = 50(1239.39 + 2260)
H = 50(3499.39)
H = 174,969.5J ≈ 175,000J
Now for the right answer to the above question:
- Option A is incorrect.
- Option B is incorrect.
- C is correct.
- D is not the correct answer.
You may please note these/this:
- Change in temperature ΔT = 100 – 80 = 20⁰C.
- This is because the water was changed to steam from 80 to 100 degrees celsius.
- We will have to convert this 20⁰C to kelvin because the specific heat capacity was given in kelvin as 4.23jg⁻¹k⁻¹ .
- The new temperature change becomes = 20 + 273 = 293k
- Always ensure consistency in units when solving scientific questions.
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/ culled from 2021 JAMB-UTME PHYSICS question 13 /