The heat of vaporization constant of water, under standard atmospheric pressure, is about 540 cal/g, that is, to vaporize each gram of water at 100 degrees Celsius (from the beginning of 'boiling') under standard atmospheric pressure, it is necessary to transfer 540 cal of thermal energy. The amount of thermal energy required to vaporize 100 g of water at 100 degrees Celsius under standard pressure is about:

Physics

To solve this question, we will follow these steps: 1. **Understand what is being asked**: The question wants to know the amount of thermal energy needed to vaporize 100 g of water at 100 degrees Celsius under standard atmospheric pressure. 2. **Analyze the given information**: The heat of vaporization constant of water is 540 cal/g. This means that for each gram of water we want to vaporize, we need 540 calories. 3. **Apply the information to the problem**: If we need 540 calories to vaporize 1 gram of water, to vaporize 100 grams we will need 100 times that value. So, we multiply 540 cal/g by 100 g. 540 cal/g × 100 g = 54000 cal 4. **Convert calories to kilocalories**: The question provides the alternatives in kilocalories (kcal), and we know that 1 kcal is equal to 1000 calories. Therefore, we need to convert 54000 cal to kcal. 54000 cal ÷ 1000 cal/kcal = 54 kcal 5. **Select the correct answer**: Based on the calculation, the amount of thermal energy required to vaporize 100 g of water at 100 degrees Celsius under standard pressure is 54 kcal. Therefore, the correct answer is alternative (d) 54 kcal.

Question about Calorimetry: Latent Heat

Subject Physics

Physics

Source URCA

URCA

Topic Calorimetry: Latent Heat

Calorimetry: Latent Heat

Difficulty Medium

Medium

(URCA 2015) - Question Medium of Physics