Question

Calculate the energy required to take 600.0 gram sample of liquid water at 30 C° is heated until half of it boils away

Answer 1

Answer: 6.69 x 10^4 joules

Step-by-step explanation:

The energy required to heat the water from its initial temperature of 30°C to its boiling point of 100°C can be calculated using the specific heat capacity of water, which is 4.18 J/g°C.

So, the energy required to heat 600.0 g of water from 30°C to 100°C can be calculated as follows:

Q1 = m x c x ΔT

Q1 = 600.0 g x 4.18 J/g°C x (100°C - 30°C)

Q1 = 150,312 J

Next, we need to calculate the energy required to boil half of the water away. The energy required to vaporize water is known as the heat of vaporization and is equal to 40.7 kJ/mol. Since one mole of water is equal to 18.02 g, the heat of vaporization for water can be calculated as 2.26 kJ/g.

So, the energy required to boil away half of the water can be calculated as follows:

Q2 = m x ΔHvap

Q2 = (600.0 g / 2) x 2.26 kJ/g

Q2 = 678.0 kJ

The total energy required is the sum of Q1 and Q2:

Total Energy = Q1 + Q2

Total Energy = 150,312 J + 678,000 J

Total Energy = 6.69 x 10^5 J

Total Energy = 6.69 x 10^4 joules.

Therefore, the energy required to take a 600.0 gram sample of liquid water at 30°C and heat it until half of it boils away is 6.69 x 10^4 joules.

Answer 2

Okay, here are the steps to calculate the energy required to heat 600 grams of liquid water at 30 C to boil away half of it:

1) Heat capacity of liquid water at 30 C is 4.18 J/kg.K. So heat capacity of 600 grams of water is 4.18 * 0.6 = 2.51 J/K.

2) To heat water from 30 C to its boiling point at 100 C requires 40 K of temperature change. So total temperature change is 40 K.

3) Energy required to heat the water = Heat capacity * Temperature change

= 2.51 J/K * 40 K

= 100.4 J

4) Latent heat of vaporization of water at 30 C is 40.7 J/g.

5) Mass of water boiled away = 300 grams (half the original mass)

6) Energy required to vaporize 300 grams of water = 40.7 J/g * 0.3 kg

= 12.21 MJ

7) Total energy required = Energy to heat the water + Energy to vaporize the water

= 100.4 J + 12.21 MJ

= 12.31 MJ

Therefore, the total energy required to heat 600 grams of liquid water at 30 C to boil away half of it is 12.31 MJ.

Let me know if you have any other questions!