Question

The initial temperature of 500 g of water is 25 degrees C. The specific heat capacity of water is 1 cal/g . degree C and the latent heat of vaporization of water is 540 cal/g.

A.) How much heat is required to raise the temperature of the water to 100 degree C?
B.) How much heat is required to vaporize the water at 100 degree C to steam?
C.) How much heat is required to turn 500 g of water initially at 25 degree C to steam at 100 degree C?

Answer 1

To turn 500 g of water initially at 25 degrees C to steam at 100 degrees C, 37,500 calories are required to heat the water to boiling, and an additional 270,000 calories are required to vaporize it, totaling 307,500 calories.

Step-by-step explanation:

The student's question involves a three-step process in turning water to steam, requiring the use of specific heat capacity and latent heat of vaporization of water. The steps are heating the water to boiling point, vaporizing it, and then adding the total heat required for both processes.

Part A: Heating Water to 100 degrees C

To calculate the heat required to raise the temperature of water to 100 degrees C from 25 degrees C, use the formula:
Heat (Q) = mass (m) x specific heat capacity (c) x change in temperature (ΔT).
For 500 g of water, this is 500 g x 1 cal/g.°C x (100°C - 25°C) = 37,500 calories.

Part B: Vaporizing Water at 100 degrees C

Using the latent heat of vaporization, the heat required to vaporize 500 g of water at 100 degrees C is:
500 g x 540 cal/g = 270,000 calories.

Part C: Total Heat Required

To turn 500 g of water at 25 degrees C to steam at 100 degrees C, the total heat required is the sum of part A and B:
37,500 calories + 270,000 calories = 307,500 calories.