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Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of

jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = $2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x$

$2330.35x=1048815\\x=450.068g$

Hence, there was 450.068g of water in the pot.

8 0

3 years ago

The physical world around us behaves as it does partly because it’s made of a huge number of tiny molecules, each behaving rando

eduard

1. Weird things like the one described above do not happen on a ramdom basis becuause molecules usually move within any enclosure in a ramdom manner. Thus, it is not possible for some types of particles to aggregate in one point while other types of molecule aggreagate in another point. Based on the kinetic energy that is available for each particle, each particle will move random
through the available space, colliding with one another and with the wall of container.

2. It will be a difficult thing to live in a Maxwell' demon world because, things will happen unpredictably and one will never know what to expect next because anything can happen at anytime. For instance, if one is drinking a glass of water, some of the particles of the water may just decide to aggregate to one part of the cup and start boiling. So, for someone who is taking a glass of water, the water may start boiling right inside his mouth when he is drinking, that will be a bad experience. When one is driving a car, the petrol particles may just decide to freeze up when one is busy speeding on the highway; that can cause a very serious accident. Thus, a world where the Maxwell law operates will be a chaotic world.

3 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$