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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.**** 17****50****00J**

**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 **

**Δ****T = 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.**

**KEY-POINTS…**

**You may please note these/this:**

**Change in temperature****Δ****T = 100 – 80 = 20****⁰****C.****This is because the water was changed to steam from 80 to 100 degrees celsius.****We will have to convert this 20****⁰****C to kelvin because the specific heat capacity was given in kelvin as 4****.23jg****⁻¹****k****⁻¹****.****The new temperature change becomes = 20 + 273 = 293k****Always ensure consistency in units when solving scientific questions.**

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**/ culled from 20****21**** JAMB-UTME ****PHYSICS**** question ****13 ****/**