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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)

A. 530000J

B. 23200J

C. 175000J

D. 130000J

QUICK ANSWER…

C

DETAILS…

Heat needed to vaporize water = heat acquired by water + heat acquired by steam

H = MCΔT + ML

H = M(CΔT + L)

Data given:

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:

1. Option A is incorrect.
2. Option B is incorrect.
3. C is correct.
4. D is not the correct answer.

KEY-POINTS…

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 /